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2.2 Chapter vx vx Velocity can be positive, t negative, or zero Acceleration can be zero in any of these cases An object with zero velocity and zero acceleration is at rest REFLECT Since acceleration is the change in velocity with time, a x SOLVE Yes Think of throwing an apple straight up in the air Establish a coordinate system in which “up” is positive On its way up, the apple’s velocity is positive While it is falling back down, the velocity is negative During the entire time the apple is in the air, the acceleration due to gravity is negative REFLECT While the apple is moving upward, it is slowing down because acceleration points downward After it has reached its highest point, acceleration still points downward, and the apple “speeds up” in the negative direction SOLVE Yes If you throw the apple of Question straight upward, it slows down to a velocity of zero at its highest point The acceleration due to gravity is still negative at this highest point REFLECT Think about a drag race with the positive x-direction toward the finish line At the instant the race starts, the vehicles have zero velocity At the same instant, the acceleration has a large positive value As a result, the velocities of the vehicles increase rapidly as they move down the track SOLVE The speed at the end of each time interval is v distance traveled t The change in speed, v v0 , is v distance traveled distance traveled1 distance traveled2 distance traveled1 t t t The change in speed with time is acceleration, so distance traveled distance traveled1 distance traveled distance traveled1 t a t t Since the time intervals are all equal, their squares are all equal We are told that the differences in the distances traveled are also equal Therefore acceleration is constant REFLECT Let us go one step further and add the distances in this series of observations After each succeeding time interval, we have traveled the distance obtained by adding all the distances traveled up through the current time interval, as shown in the table below Number of time interval Total distance traveled 00 1 1 1 1 16 25 With an initial speed v0 = 0, these values match the results of the kinematic equation for distance with constant acceleration: distance traveled 21 a t where a 2m s2 and t increases by one second for each time interval Motion in One Dimension 2.3 SOLVE REFLECT Look at a point on a velocity-versus-time graph This point gives the slope of the line on the corresponding position-versus-time graph at that same time The slope of the velocity-versus-time graph gives the location of a point on the acceleration-versus-time graph at that same time SOLVE Suppose your car has a maximum speed of 100mi h, or about 45m s At an acceleration of 3.0m s , it 45 m s would take t 15 s to reach that maximum speed 3.0 m s2 REFLECT This answer assumes that this maximum acceleration remains constant In practice, acceleration depends on the gear you are in and the speed of the engine (we’ll study more about what is known as “torque” in Chapter After the transmission shifts into higher gears, the acceleration decreases As you speed up, air resistance also increases This also contributes to decreasing the acceleration SOLVE The velocities of each of the two cars are given as constant, with no mention of change in velocity The acceleration of each car must be zero REFLECT If the problem had stated that the two velocities were given only for a specific point in time, then the answer might have been ambiguous An alternative is that the faster car has zero acceleration while the slower car has a positive acceleration The faster car passes the slower car, but in a short time, the car that was slower will accelerate to a speed faster than 25 m s and will pass the first car This is what happens when you pass a stationary police car while you are driving at constant speed over the speed limit! 10 SOLVE At a given position, velocity is different, speed is the same, and acceleration is the same REFLECT At any particular position or height, velocity has the same magnitude, but opposite sign Going up, the sign on velocity is positive Coming back down, the sign is negative Speed does not change because v vy , so the sign of v y does not matter Acceleration is constant during free fall, with ay g 9.80 m s 11 SOLVE If an object’s average velocity is zero, then its displacement must be zero x vx t For vx 0, x t m If the object’s acceleration is zero, then the displacement might be zero, positive, or negative, depending on the value and sign of the velocity REFLECT Even if the object’s instantaneous velocity has not been zero throughout the time interval, a velocity of zero implies that it has returned to its starting point, so x If acceleration is zero and velocity is zero, then the 2.4 Chapter object is at rest and has not moved If velocity is positive or negative, then displacement constantly increases in the same direction as velocity 12 SOLVE With a sufficient number of steps, you will eventually reach your destination If we graph displacement versus time, we see that even though we divide the distance traveled by two in each successive step, we also divide the time it takes by two The speed therefore remains constant Therefore, if you cover half the distance to your goal in 21 t, you will also cover the remaining half in an additional time of t, no matter the number of parts into which it is divided REFLECT In fact, the distance is divided into an infinite number of steps Mathematicians have found this to be a “summable” infinite series, written as 1 1 16 32 1 2n So the sum of this infinite number of steps toward the goal is simply the distance to the goal Since the paradox gives no limit on the number of steps and velocity remains constant, you can reach your goal in exactly twice the time it takes you to go half the distance toward the goal MULTIPLE-CHOICE PROBLEMS 13 ORGANIZE AND PLAN We are to find average speed We use the definition of average speed, v x t The responses are all in m s, so we must convert distance to meters and time to seconds Known: x 385,000 km; t 2.5 day SOLVE Using the definition of speed, vx x t vx 385,000 km 1000 m km 1800 m s 2.5 day 24 h day 3600 s h The answer is response (C) REFLECT The escape velocity from Earth is about 11,000 m s A spacecraft traveling toward the moon initially achieves escape velocity, and then slows down due to Earth’s gravity If the intention is to make a “soft” landing on the moon with vx m s, it makes sense that the average speed must be less than the escape velocity 14 ORGANIZE AND PLAN We are given time and speed and are asked to find distance We use the definition of speed, v x t and rearrange it to solve for distance Known: v 32 m s; t 35 s SOLVE The definition of speed is v x t Rearranging, we get x v t Substituting known values, x 32 m s 35 s 1100 m The answer is response (D) REFLECT Even though the problem states that the cheetah’s top speed is 32 m s we consider it to be the average speed for the duration of this problem A distance of 1100 m is about 1200 yards, the length of 12 football fields This is a reasonable distance for a cheetah to chase its prey 15 ORGANIZE AND PLAN In this problem, the runner has already run part of the race with known quantities We have to figure out what’s going to happen during the rest of the race We’ll use subscript “1” for the first part of the race, subscript “2” for the remainder of the race, and no subscript if the variable applies to the entire race We need to find the speed v2 the runner must maintain until the finish line We know that it’s a 1500 m race, with 1200 m Motion in One Dimension 2.5 already run Since the runner must finish the race in under minutes, our strategy is to find the time already elapsed and then the allowed time remaining From that, we can calculate the necessary speed Known: x 1500 m; x1 1200 m; v1 6.14 m s; t SOLVE For the part of the race already completed, we use x v1 t1 Solving for t1 , we get t1 x1 1200 m 195.44 s v1 6.14 m s t2 min 60 s 195.44 s 44.56 s x2 1500 m 1200 m 300 m v2 x2 300 m 6.73 m s t2 44.56 s The answer is response (a) REFLECT This is nearly a 1-mile race It is normal for runners to sprint, or run faster, at the end of a long race The required average speed to finish under minutes is only a little faster than the average speed in the first part of the race, which is reasonable 16 ORGANIZE AND PLAN In this problem, the runner travels forward and backward along the x-axis We are asked to calculate the runner’s average velocity We can’t just average the two velocities given; this would violate math rules for the definition of velocity Our strategy is to calculate the total displacement and the total time From these we can calculate the average velocity for the entire run We’ll use subscript “1” for running forward, subscript “2” when jogging backward, and no subscript for a variable that applies to the entire run Since we are calculating average velocity rather than speed, we carefully note the signs of the values Known: vx1 9.2 m s; vx 3.6 m s; x1 100 m; x2 50 m SOLVE First, the total displacement x x1 x2 100 m 50 m 50 m The runner ends up only 50 m from the starting point Next, the total time x 100 m t1 10.87 s v x1 9.2 m s t x2 50 m 13.89 s v x 3.6 m s t t1 t2 10.87 s 13.89 s 24.76 s Finally, the average velocity vx x 50 m 2.0 m s t 24.76 s The answer is response (a) REFLECT Velocity is based upon displacement, not distance The farther backward (toward the origin) the runner jogs, the smaller the displacement and the closer the average velocity will approach zero 17 ORGANIZE AND PLAN This problem is about the properties of displacement in one-dimensional motion and comparison of displacement with distance SOLVE The definition of displacement in one-dimensional motion is the difference between final and initial positions in a coordinate system: x x x0 2.6 Chapter A one-dimensional coordinate system extends without bound in both the positive and negative directions from the origin Therefore both x and x0 can have values that are positive, negative, or zero, independently of each other Since this is the case, the difference between the two values can also be positive, negative, or zero The answer is response (b) REFLECT Distance is the sum of the lengths of all the straight-line segments of a trip, without regard to sign Distance is always positive The displacement value may be equal to the distance value, but does not have to be Displacement is never greater than distance 18 ORGANIZE AND PLAN In this problem we are to find average speed from two different speeds and distances run at those speeds We don’t just average the speeds We must divide the total distance by the total time, so we have to find each of these first We’ll use the formula for average speed v x t Known: v1 4.0 m s; v2 6.0 m s; x1 x2 60 m SOLVE The total distance is just the sum of the two distances x x1 x2 60 m 60 m 120 m Now we find the times for each distance run, using the formula for speed t1 x1 60 m 15 s v1 4.0 m s t2 x2 60 m 10 s v2 6.0 m s t t1 t2 15 s 10 s 25 s v x 120 m 4.8 m s t 25 s The answer is response (a) REFLECT The answer is an average The value we found lies between the higher and lower values, as it should However, note that the answer is not just the arithmetic mean of the two given velocity values 19 ORGANIZE AND PLAN We must find average acceleration from initial velocity, final velocity, and elapsed time This means we use the definition of average acceleration, ax vx t Known: vx 1250 m s; vx 1870 m s; t 35 s SOLVE We use the definition of average acceleration ax vx vx vx 1870 m s 1250 m s 17.7 m s t t 35 s The answer is response (c) REFLECT The acceleration of the spacecraft may seem modest, about a 50% increase in velocity But the velocity values themselves are not important Only their difference matters, vx In this case the acceleration is about 180% of that due to gravity If this spacecraft has human occupants, acceleration of this magnitude would certainly be noticeable By comparison, a sports car accelerating from zero to 60 miles per hour in seconds only accelerates at about 68% of that due to gravity 20 ORGANIZE AND PLAN This problem requires us to find displacement from initial velocity, final velocity, and acceleration Known: a x 1.4 m s ; vx 10 m s; vx0 m s SOLVE Since we don’t know time, we will make use of the kinematic equation v2x v x20 2a x x Rearranging to find x , v x2 v x20 10 m s m s 35.7 m x 2a x 1.4 m s 2 The answer is response (c) Motion in One Dimension 2.7 REFLECT Note that squaring the velocities results in a positive numerator The sign of the acceleration is negative, producing a negative displacement, as we would expect from an object starting from rest 21 ORGANIZE AND PLAN This problem requires us to find acceleration from initial velocity, final velocity, and displacement Known: vx0 21.4 m s; vx m s x 3.75 cm SOLVE Since we don’t know time, we will make use of the kinematic equation v2x v x20 2a x x Since x is given in centimeters, we need to convert it to meters Rearranging to find ax , m s 21.4 m s 6100 m s2 v2x v2x ax 2x 3.75 cm 1 m 100 cm 2 The answer is response (c) REFLECT The arrow has a relatively high speed and stops in a very short distance This is like an automobile traveling 50 mi/h stopping in about 1.5 inches This requires a very large negative acceleration The sign on the acceleration is negative because the initial velocity of the arrow is positive, and it slows down as it travels into the target 22 ORGANIZE AND PLAN This is a free-fall problem, with the object in constant acceleration We want to know how long it takes to strike the ground, in seconds We will have to use one of the kinematic equations that contain time as a variable Since we know that initial velocity is zero, but we don’t know final velocity, we will use y vy0 t 21 a y t and solve for t Known: y 442 m; ay g 9.81 m s2 ; vy m s SOLVE We start with the formula we chose y vy0 t 21 a y t Since v y 0, y 21 ay t 2y t ay t 2y ay 441 m 9.80 m s 9.5 s The answer is response (c) REFLECT The Sears Tower is one of the world’s tallest buildings It takes an object less than 10 seconds to fall to the ground, ignoring air resistance This emphasizes the important aspect of constant acceleration The longer the object falls, the faster it goes, and the greater the distance it covers each second We see that the height of the building appears under the radical Doubling the height of the building will not double the length of time the object takes to fall from the top Rather, the time increases by a factor of 23 ORGANIZE AND PLAN This is a free-fall problem Notice that only an algebraic solution is needed We need to find the effect on final velocity if we double the distance an object falls We won’t use a subscript for the original height, but we’ll use the subscript “2” for the doubled height Since time is not included in the problem, we’ll use the kinematic equation vy2 vy20 2ay y Known: y h; v y m s; vy v; a y g SOLVE Substituting the variables given in the problem, we get v2 v02 2a yh 2.8 Chapter Since v y for both the original height and the doubled height, v2 2ah v ah If we now double y to the value 2h, then the new velocity is v2 2a y 2h 2ah v The answer is response (c) REFLECT Displacement in the y-direction is a function of the square of velocity It makes sense that velocity is a function of the square root of height during free-fall 24 ORGANIZE AND PLAN This is a constant-acceleration problem Since time is not part of the problem, we can use the kinematic formula v 2x v x20 2a x x Known: vx m s; vx 12.8 m s; x 16.0 m SOLVE Starting with our chosen formula and solving for acceleration, v2x v x20 2a x x ax v2x vx20 2x Substituting known values, ax m s 12.8 m s 5.1 m s 2 16.0 m The answer is response (b) REFLECT We can think of this problem as the opposite of a free-fall problem (and in the horizontal direction) Since the car initially has positive velocity and is slowing down, the acceleration must be negative 25 ORGANIZE AND PLAN We are asked to find the shape of the graph of velocity versus time for a moving object under constant acceleration Since the object is moving, we know that velocity can be positive or negative, but it cannot be zero No calculation is needed Known: v m s; a x m s SOLVE We know that velocity is a function of time and can be expressed as vx vx ax t We rearrange this formula to put it in the slope-intercept form of the equation of a straight line vx ax t vx We see that ax is the slope and vx is the vertical intercept (the intercept on the velocity axis) We are told that ax The slope of the line must be either positive, sloping upward, or negative, sloping downward Either of these possibilities gives a diagonal line on the graph The answer is response (b) REFLECT An acceleration of zero would give us a horizontal line (slope = zero) In order for the graph to have the shape of a parabola, velocity would have to be a function of t Rather, it is a function of the first power of t , a linear function 26 ORGANIZE AND PLAN This problem contains two objects! We must find some common value between them In this case, both the time at which they pass and their height y as they pass one another are common We solve for the height at which each ball passes the other and set the two equations equal to each other From this, we find a numeric value for t Substituting this value back into the equation for height for either ball or ball gives us the value for height Known: vy1 10m s; ay g 9.80 m s2 ; v y 10 m s; y2 10 m SOLVE The two balls not only pass each other at the same height, the must pass each other at the same time Our key is to find that time For the first ball, y y1 vy1t 21 a y t Motion in One Dimension 2.9 For the second ball, y y2 vy2 t 21 ay t Since the balls meet at the same height y, we can set these two equations equal: y1 vy1t 21 ay t y2 vy2t 21 a y t Canceling like terms and substituting known values, we get 10 m s t 10 m+ 10 t m s t 10 m 0.5 s 20 m s We know that the height is the same for both balls, so we choose the equation for the position of the first ball Substituting known values including the time, y 10 m s 0.5 s+ 12 9.80 m s 0.5 s 3.8 m The answer is response (c) REFLECT The two balls have to meet somewhere between ground level and 10 m above the ground, which they Checking our work with the equation for the position of the second ball, y 10 m 10 m / s 0.5 s 9.80 m / s 0.5 s y 3.8m PROBLEMS 27 ORGANIZE AND PLAN In this problem we must show the difference between the distance and the displacement for a round trip between two points We’ll use “1” as the subscript for the first part of the trip and “2” as the subscript for the second part of the trip Known: d1 200 m SOLVE The distance between two points is always positive regardless of the direction traveled For a round trip to the video store, the distance from your friend’s house is d1 200 m The distance from the video store back to your friend’s house is also d2 200 m So the total distance for the round trip is d d1 d2 200 m 200 m 400 m But for displacement, we must take into account the sign of the direction of travel for each part of the trip Traveling in the positive direction, from your friend’s house to the video store, x1 200 m Returning from the video store in the negative direction, x2 200 m The total displacement for the round trip is x x1 x2 200 m 200 m m REFLECT A round trip, with the ending position the same as the starting position, always has a positive distance and a zero displacement 28 ORGANIZE AND PLAN This problem uses the definition of displacement We choose a coordinate system with the positive direction to the east Since Lincoln is our starting point, we’ll choose the position of Lincoln to be x0 Our final position will be Grand Island, 160 km to the west Its position is x Known: x0 km; x 160 km SOLVE We travel in the negative direction from Lincoln to Grand Island Using the definition of displacement, x x x0 160 km km 160 km 2.10 Chapter REFLECT When you give someone directions to a destination, you must tell them which direction to travel if you want them to arrive where they intend to go When traveling in the negative direction, the value of displacement will have a negative sign 29 ORGANIZE AND PLAN In this problem, we have to consider what effect a fractional round trip has on distance and displacement We set up a coordinate system with the positive direction to the east We start at Grand Island, so we can declare this position to be x0 The position of Lincoln will be x SOLVE (a) Here we have three round trips from Grand Island to Lincoln The displacement x1 traveling from Grand Island to Lincoln is x1 x x0 160 km km 160 km Traveling from Lincoln back to Grand Island, the displacement is x2 160 km The total displacement for one round trip is x 160 km 160 km km Therefore, the displacement for three round trips is 3x km km The distance traveled from Grand Island to Lincoln is 160 km Since distance is always positive, the distance traveled from Lincoln back to Grand Island is also 160 km The distance traveled in three round trips is d 160 km 160 km 960 km (b) Here we have 21 round trips This means three round trips plus one last trip from Grand Island to Lincoln Since the displacement from (a) for exactly three round trips was km, the displacement here is the same as onehalf of a round trip, or the displacement from Grand Island to Lincoln, so x x x0 160 km km 160 km However, the distance for 21 round trips from Grand Island to Lincoln is d 160 km 160 km 160 km 160 km 1120 km (c) This part asks about displacement and distance for 34 round trips From (b), the displacement for round trips is zero During the remaining 34 round trip, we travel from Grand Island to Lincoln in the positive direction, and then halfway back to Grand Island in the negative direction The displacement for this trip is x 160 km 12 160 km 80 km The distance for each leg of the trip is always positive, so d 160 km 12 160 km 1200 km REFLECT The displacement for a round trip is always zero The distance for each leg of a trip is positive, regardless of direction Think about how often a car’s owner must fill the fuel tank just from driving round trips to school! 30 ORGANIZE AND PLAN We must find the average speed in m s of a runner under three different conditions of time and distance For each of these, we’ll use the definition of average speed, v distance traveled However, in each t scenario, we’ll have to convert time or distance, or both, to the proper units Known: (a) distance traveled 41 km; t h 25 min; (b) distance traveled 1500 m; t 50 s; (c) distance traveled 100 m; t 10.4 s SOLVE (a) Start by converting h 25 to seconds We think of this as h 25 and convert each term separately: t h 3600 s h 25 60 s 8700 s Motion in One Dimension 2.11 Then we convert 41 km to m, distance traveled 41 km 1000 m km 41,000 km Now we use the definition of average speed distance traveled 41,000 m v 4.7 m s t 8700 s (b) The distance is already given in meters We convert the time to seconds t 50 s 60 s 50 s 230 s Then v distance traveled 1500 m 6.5 m s t 230 s (c) Here the distance is already in meters and the time is already in seconds distance traveled 100.m v 9.62 m s t 10.4 s REFLECT The lengths of these races systematically decrease from very long (a marathon!) to moderate (just under a mile) and finally to a short dash We can expect the average speeds to increase from a bit more than a fast walk ( 4.7 m s ) to a fast run ( 9.62 m s ) 31 ORGANIZE AND PLAN To calculate elapsed time, we need to know average distance between the Earth and the sun, and the speed of light We use 3.00 108 m s for the speed of light and 1.50 1011 m for the orbital radius of Earth Known: distance traveled 1.50 1011 m; v 3.00 108 m s SOLVE Using the definition of average speed, distance traveled v t and rearranging for t , t distance traveled 1.50 1011 m 500 s v 3.00 108 m s REFLECT One occasionally hears that if the sun suddenly stopped shining, it would take the inhabitants of Earth about minutes to realize this Checking this value, 500 s 8.33 or about minutes and 20 seconds 60 s 32 ORGANIZE AND PLAN This problem requires us to rearrange the definition of average speed, solving for time Then we are given two instances in which we must find time from speed and distance Known: (A) distance traveled 18.4 m; v 44 m s; (B) distance traveled 18.4 m; v 32 m s SOLVE Rearranging the definition of average speed we obtain t distance traveled v (a) t 18.4 m 0.42 s 44 m s (b) t 18.8 m 0.58 s 32 m s REFLECT Major league batters have about one-half second to see, track, and hit a pitch The faster the pitch, the less time the batter has 33 ORGANIZE AND PLAN In this problem there are two parts to the motion, at different speeds We not simply average the speeds, even though the distances traveled are both 100.m We must go back to the definition of Motion in One Dimension 2.39 Since v2 v0 m s, a1y a2 y sin 1 sin Canceling and rearranging for a2 , sin2 sin 45 a2 a1 3.50 m s 4.95 m s sin sin30 1 REFLECT The magnitude of the acceleration is greater rolling up the steeper slope It is also negative, because the ball is slowing down In this problem, neither the initial nor final motion is parallel to either the x- or y-axis Since the angles aren’t complementary, there’s no advantage to tilting the axes as in Figure 2.2 If it appears that there is not enough information given to find the answer to a problem, the best strategy is to set the problem up and solve it algebraically before trying to substitute known values The unknown variables will cancel, such as v0 , v2 , and y did in this problem You will be left with a simplified and more easily solvable problem 86 ORGANIZE AND PLAN This is a free-fall problem involving a wrench, but we only know a y and y We need one more variable to solve this problem We can calculate the initial speed of the wrench from the motion of the helicopter The helicopter is in constant acceleration from rest on the ground We know the helicopter’s initial speed, displacement and acceleration and can calculate its speed at y 20 m Since the wrench is in the helicopter, the initial speed of the wrench as it falls out is the same as the final velocity of the helicopter Now we can use v v02 a y y to find the final speed Then we can use the definition of acceleration to find time 2 Known: vh m s; ayh 0.40 m s ; ayw g 9.80 m s ; yh 20 m; yw 20 m SOLVE First, find the upward speed of the helicopter (and hence the wrench) at 20.0 m 2 vyh vyh a yh y 0 vh vh20 a yh y m s 0.40 m s 2 20 m 4.0 m s Next, use the velocity of the helicopter to find the final velocity of the wrench as it strikes the ground vw0 vh 4.0 m s vw vw0 2 2aywy The wrench is falling downward, but the problem only asks for speed, so vw 4.0 m s 9.80 m s 20 m 20.2 m s 20 m s Now, using the definition of acceleration, t w vw vw 20.2 m s 4.0 m s 2.5 s ayw 9.80 m s (b) For the wrench’s free fall to the ground, from (a) vw 20 m s REFLECT The wrench’s initial velocity is positive and its final velocity is negative, as we would expect The time to fall 20.0 m is reasonably between and seconds It is a math coincidence that the magnitude of vyw and yw are the same This has no special meaning and is due to the choice of values in the problem 87 ORGANIZE AND PLAN Now that we know the initial velocity and the time to impact of the wrench, we can graph position versus time and velocity versus time Known: yw0 20.m; vyw0 4.0 m s; a yw g 9.80 m s SOLVE To find ordered pairs of position and time, we use yw yw0 vyw0 t 21 a yw t between t and t 2.5 s To find ordered pairs of time and velocity, we use vyw vyw0 ayw t 2.40 Chapter We construct a table with (arbitrarily) 12 ordered pairs: t ,s yw ,m vyw ,m s 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.5 20.0 20.6 20.8 20.6 20.1 19.1 17.7 16.0 13.9 11.3 8.4 5.1 4.0 2.0 0.1 1.9 3.9 5.8 7.8 9.7 11.7 13.6 15.6 17.6 20.2 Now we can plot position versus time, shown in the first figure The graph of velocity versus time is shown in the second figure REFLECT It makes sense that the wrench continues upward for a short time and then its velocity becomes negative, responding to the acceleration of gravity We notice from the table and graphs that the wrench seems to “hang” in the air near its highest point, like the volleyball player in Problem 81 88 ORGANIZE AND PLAN This is a constant acceleration problem, since the flea is in free fall after it leaves the ground We know its displacement, final velocity and acceleration, so we can solve the problem We’ll use Motion in One Dimension v2y vy20 2ay y to find initial velocity Then we’ll use the definition of acceleration a y vy vy0 t 2.41 to find the time to reach its final height Known: For the flea ascending though the air: vy m s; a y g 9.80 m s ; y 30.cm For the flea during the act of jumping: vy m s; y 0.90 mm SOLVE (a) First we convert the height to meters 1m y 30.cm 0.30 m 100 cm Then we find the takeoff speed, vy v2y vy20 2ay y Since vy 0, vy 2a y y 2 9.80 m s2 0.30 m 2.42 m s (b) Then we find the time the flea takes to reach 0.30 m t vy vy0 ay m s 2.42 m s 9.80 m s2 0.25 s (c) When the flea is in the process of jumping into the air, 1m 4 y 0.90 mm 9.0 10 m 1000 m v2y vy20 2ay y Since vy m s , v2y 2a y y and ay v2y 2y 2.42 m s 2 9.0 104 m 3300 m s2 This acceleration is about 330 times g REFLECT The velocity of the flea in the air is positive and a y is negative, so the flea slows down as we expect The acceleration of the flea is very large as it launches itself into the air There is a great difference between a human brain and the nervous system of a flea (see Problem 72) 89 ORGANIZE AND PLAN We’ll find the final speed of a cat falling through a distance of 6.4 m We know displacement, initial speed, and acceleration, so we’ll use v v02 2ay y Then we’ll use the same formula with the final velocity we just calculated to find the acceleration of the cat as it crouches to a stop Known: When the cat is in free fall: v0 m s; y 6.4 m; ay g 9.80 m s As the cat crouches: y 14 cm SOLVE (a) To find final velocity in free fall, vy2 vy20 2a y y Since v0 m s, v a y y 9.80 m s 6.4 m 11 m s (b) First convert y to meters: 1m y 14 cm 0.14 m 100 cm Then as the cat crouches, v v02 a y y 2.42 Chapter Using unrounded values, v v02 m s 11.2 m s 448 m s y 0.14 m ay REFLECT 90 Cats can endure greater acceleration than humans (see Problem 72) ORGANIZE AND PLAN The motion in this problem is part constant acceleration and part constant velocity We’ll divide the problem into these two parts to solve We’ll use the subscripts and for the two parts During constant acceleration we use vx vx ax t to find final velocity To find distance, we’ll use x vx t 21 a x t During the constant velocity portion, we’ll use x vx t to find displacement Known: Constant acceleration: vxo m s; ax 2.0 m s2 ; t 5.0 s Constant velocity: t 5.0 s SOLVE (a) Your final velocity is reached at the end of the period of constant acceleration: vx vx ax t m s 2.0 m s2 5.0 s 10.m s (b) The displacement during constant acceleration is x1 vx 0t 21 ax t m s 21 2.0 m s2 5.0 s 25 m 2 During constant velocity, x2 vx t 10.m s 5.0 s 50.m xtotal x1 x2 25 m 50.m 75 m (c) We complete a table from which we can select ordered pairs t ,s x,m vx ,m s 10 16 25 35 45 55 65 75 10 10 10 10 10 10 We graph ordered pairs t, x as shown in the figure below Motion in One Dimension 2.43 Then we graph ordered pairs t , v x as shown in the figure below REFLECT Your displacement increases faster and faster until you reach t 5.0 s Then it increases linearly with time Velocity increases linearly with time until you reach t 5.0 s Then it “levels off ” to vx 10.m s 91 ORGANIZE AND PLAN Here we know initial and final velocity and displacement We need to find acceleration The problem is independent of time, so we use vx2 v x20 2a x x Once we have found ax we find time using the vx vx In the second part of the problem, we have a constant-velocity period t before we accelerate For this period we use vx vx ax t to find the displacement, x1 Finally we find the definition of acceleration, a x acceleration needed to stop in the remaining distance, x2 Known: (a) and (b) vx0 13.4 m s ; vx m s; x 15.0 m (c) Constant velocity: t 0.60 s; vx m s SOLVE (a) Rearranging for ax , v2x v x20 2a x x v2x v2x m s 13.4 m s 5.99 m s2 2x 15.0 m ax (b) Rearranging for t , ax vx vx t Using the unrounded value for ax t v x v x 0 m s 13.4 m s 2.24 s ax 5.98 m s2 (c) First we find how far we went in the 0.6 s reaction time: x1 vx t 13.4 m s 0.6 s 8.04 m This leaves the distance in which we have to stop as x2 15.0 m x1 15.0 m 8.04 m 6.96 m Using the same formula as in (a), v2x v2x m s 13.4 m s 13 m s2 2x 6.96 m ax REFLECT With a positive velocity, we expect negative acceleration in order to slow down The acceleration in (c) is a rather rapid acceleration, about one-third greater than that of gravity 92 ORGANIZE AND PLAN A ball is projected up an incline We are to describe the motion of the ball We’ll use x 21 vx vx t to find how far up the incline the ball reaches before rolling back down Finally, we’ll draw two graphs showing acceleration versus time and position versus time for the rolling ball 2.44 Chapter Known: (a) t 4.0 s; vx 1.0 m s; vx 1.0 m s; (b) t 2.0 s; vx 1.0 m s; vx m s; SOLVE (a) The ball will start up the incline at 1.0 m s, accelerate to a stop, and then roll back down, accelerating to 1.0 m s It would look like “free fall at an angle.” (b) We find the distance up the ramp at the point where vx m s and t 2.0 s from the figure above x vx vx0 t 21 m s 1.0 m s 2.0 s 1.0 m (c) We see from the figure above that ax is constant We calculate ax vx vx 1.0 m s 1.0 m s 0.5 m s2 t 4.0 s and plot the line ax 0.50 m s2 , which appears as a horizontal line on our graph, shown in the figure below Now we’ll need to construct a table in order to find ordered pairs t, x to plot position versus time We’ll get our values of x from x 12 vx vx t We’ll calculate a value for every 0.50 s increment since these values are directly readable from the figure in part (a) t ,s 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 x,m 0.0 0.4 0.8 0.9 1.0 0.9 0.8 0.4 0.0 Motion in One Dimension 2.45 Then we plot the ordered pairs t, x Our graph is shown in the figure below REFLECT This problem is just like a free-fall problem except the value for acceleration is less than g In Chapter 3, we’ll learn how to use and evaluate x- and y-components of a value that has a direction, such as “up the incline.” 93 ORGANIZE AND PLAN We know the values of three variables in this constant acceleration problem In order to graph velocity and position versus time, we need to find acceleration We’ll use the definition of acceleration, ax vx vx Once we know ax we can construct a table of values to graph, using v2x vx20 2a x x to find t position Known: vx 25 m s; vx m s; t 10.0 s SOLVE (a) Using the definition of acceleration, ax vx vx 0 m s 25 m s 2.5 m s2 t 10.0 s Rearranging the definition of acceleration to find velocity, vx vx ax t Calculating vx at 2.0-s intervals, t, s vx.m/s 2.0 4.0 6.0 8.0 10.0 25 20 15 10 5.0 Plotting ordered pairs of t , v x we get the graph shown in the figure below 2.46 Chapter (b) Finding values of x at the end of each 2.0-s time interval using x v x 0t 12 a x t , x t, s 2.0 4.0 6.0 8.0 10 45 80 105 120 125 Plotting ordered pairs of t , x we get the graph shown in the figure below (c) A motion diagram shows position at equal time intervals We can use the data from the table in (a) to this: REFLECT The slope of the graph in (b) represents the acceleration The slope is downward and to the right, corresponding to a negative acceleration Looking at the graph in (c), we see that forward motion continues ever slower, until the vehicle stops 94 ORGANIZE AND PLAN In order to graph the motion of the ball, we need an equation that describes its motion We need to use the value of acceleration in order to calculate velocity and position at each point in time, so we’ll calculate ax first We know time, displacement, and initial velocity, so we can solve for displacement using x v x 0t 12 a x t For velocity, we’ll use vx vxo ax t We take the x-axis to be parallel to the incline Known: vx 2.40 m s ; t 6.0 s x m SOLVE (a) First, we declare the initial position to be zero x0 m x vx t 21 a x t Since x 0, t v x 21 a x t Setting the second factor equal to zero, vx 21 ax t ax 2vx 2 2.40 m s 0.80 m s2 t 6.0 s Knowing ax , we can complete a table showing values of vx and x as functions of time We’ll use time intervals of 0.50 s Motion in One Dimension t ,s vx x,m 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00 2.40 2.00 1.60 1.20 0.80 0.40 0.00 –0.40 –0.80 –1.20 –1.60 –2.00 –2.40 1.10 2.00 2.70 3.20 3.60 3.60 3.50 3.20 2.70 2.00 1.10 2.47 Graphing ordered pairs of velocity versus time t , v x we get the graph shown in the figure below (b) Then graphing ordered pairs of position versus time t , x we get the graph shown in the figure below (c) Finding average velocity, vx x m 0 m s t 6.0 s Then we find the total distance traveled The distance to the highest point on the incline is 3.60 m, so distance traveled1 3.60 m Likewise, the distance from the highest point back down to the point of beginning is distance traveled2 3.60 m Total distance traveled is distance traveledtotal distance traveled1 distance traveled2 3.60 m 3.60 m 7.20 m 2.48 Chapter REFLECT The graph of velocity versus time is a straight line, as we expect from the linear relationship vx vx ax t It has a negative slope, reflecting the negative value of a x The graph of position versus time is a parabola, as we expect from the term x contained in the formula for position 95 ORGANIZE AND PLAN In this problem we’re asked to find the time(s) when the ball is 5.20 m above its launch point Those parentheses are important! The only case in which there would be only one time is if the ball is at its maximum height at that time We can test this by using the quadratic formula to see if we get two positive roots, which are solutions for time After that, we can use v2x vx20 2a x x to find the velocities at each of these points in time Known: vy 12.1 m s; a y g 9.80 m s ; y 5.20 m SOLVE (a) We start with x vx t 21 a x t Rearranging to the general form of a quadratic, a y t 2 vy0t y Substituting the coefficients into the quadratic formula, t vy vy0 21 ay y 21 a y Simplifying, t vy vy0 2a y y ay Substituting known values, t 12.1 m s 12.1 m s 2 9.80 m s2 5.20 m 9.80 m s2 The roots of this equation are 12.1 6.67 t m s 0.554 s and 9.80 12.1 6.67 t m s 1.92 s 9.80 (b) To find the final velocity, we use v2y vy20 2ay y v y v y20 a y y 12.1 m s 2 9.80 m s 5.20 m 6.67 m s REFLECT We notice that the expression for final velocity also appears in the numerator of the quadratic formula We can deduce that on the way back down, the ball will strike the ground at 1.915 s 0.554 s 2.47 s The 0.554 s is the same time interval it took for the ball to reach a height of 5.20 m on the way up 96 ORGANIZE AND PLAN This problem is similar to Problem 80 except we have no evidence that the upward and downward velocities have the same magnitude It would be a coincidence if they did, so all we know is that the position and time are the same for both balls We’ll solve for position in terms of time for each ball Then we’ll set the two equations equal and solve for time Once we know time, we can solve for the position of one ball, which will be the same for the other ball We’ll use subscripts and to identify the variables that apply to each ball Known: Ball 1: vy 11.0 m s; a y g 9.80 m s ; y1 1.50 m Ball 2: vy 11.0 m s; a y g 9.80 m s ; y2 12.6 m Motion in One Dimension 2.49 SOLVE First, the position of ball y y01 v y 01t 21 a y t For ball 2, y y02 vy 02 t 21 a y t Combining these two equations, y01 vy 01t 21 a y t y02 vy 02t 21 a y t 2 Simplifying, y01 y02 vy 02 t vy01t and t 1.50 m 12.6 m 0.505 s y01 y02 vy02 vy01 11.0 m s 11.0 m s Substituting the value of t into the equation of position for ball 1, y 1.50 m 11.0 m s 0.505 s 21 9.80 m s2 0.505 s 5.81 m The two balls meet at a height of 5.81 m at 0.505 s after they are thrown REFLECT Examining the solution, we see that this is just like assuming that one ball is standing still and both ball and the surroundings are moving 97 ORGANIZE AND PLAN Here we have to describe the motions of the friends in a frame of reference that is moving with respect to the “stationary” fixed ends of the “moving sidewalk.” We’ll use subscripts and for the friends, and for the sidewalk Known: vx3 1.0 m s with respect to the fixed ends; vx1 4.00 m s with respect to the sidewalk; vx 4.0 m s with respect to the sidewalk SOLVE From the frame of reference of the sidewalk, each of the two friends, traveling at the same speed, will travel 25 m, meeting in the middle of the sidewalk For each person, x vx t t x 25 m 6.25 s t 4.0 m s In the same time period, the entire frame of reference of the sidewalk has moved x3 vx 3t 1.0 m s 6.25 s 6.25 m So in the frame of reference of the fixed ends of the sidewalk, the friends meet at 25 m 6.25 m 31 m from one of the fixed ends An alternative solution is to consider the relative velocities of the two friends with respect to the fixed ends The first person will be traveling 5.0 m s while the second person will be traveling 3.0 m s x1 5.0 m s t x2 3.0 m s t x1 x 50.m t 5.0 m s 3.0 m s x1 5.0 m s 50.m 31.25 m 31 m 8.0 m s So the answer is 31 m from one end REFLECT This is an example of relationships between the motions of objects compared in two frames of references that are moving at constant velocity with respect to each other 2.50 98 Chapter ORGANIZE AND PLAN In this problem, we have to compare the position of the gazelle with the position of the cheetah at the moment the cheetah would tire, after which the cheetah would accelerate to a stop If the cheetah has not caught the gazelle during this time interval, then the cheetah will not catch the gazelle at all The gazelle has a constant velocity and a head start of 25 m, so we use vx vx ax t The cheetah accelerates to a constant velocity, so for the cheetah, we can use a combination of x vx t 21 a x t and the equation we used for the gazelle We’ll use the subscripts c and g to represent the cheetah and the gazelle, and subscripts a and v for periods of constant acceleration and velocity Known: x0 g 25 m; v xg 20.m s; xc m; vx0c m s; vxc 60.mi h; during cheetah’s constant acceleration: ta 3.0 s; during cheetah’s constant velocity: tv 10.s SOLVE (a) First we convert the speed of the cheetah to SI units: 5280 ft 0.3048 m h v xc 60.mi h 27 m s mi ft 3600 s Then we calculate the distance traveled by the cheetah during constant acceleration: v2x v x20 2a x x Since vx0c m s, 26.8 m s 40.m v2x 2a y 8.93 m s2 xca During constant velocity, the cheetah runs xcv vxc t 26.8 m s 10.s 268 m The cheetah can run a total of xc xca xcv 40.2 m 268 m 310 m (b) The total time of this event is ttotal ta tv 3.0 s 10.0 s 13 s At the end of this time interval, the position of the gazelle is xg x0g v xg t total 25 m 20.m s 13 s 285 m Since the gazelle can only travels 285 m in the time it takes the cheetah to travel 310 m, yes, the cheetah catches the gazelle before tiring REFLECT We see that the motion of each animal is stated separately as if the other did not exist If we were to calculate the time at which the cheetah catches the gazelle, we would see that it’s likely the cheetah doesn’t actually run the entire 10 s, but stops for a meal instead 99 ORGANIZE AND PLAN The runners meet at the same time, t We can consider this a frame-of-reference problem Suppose that Runner A is standing still and Runner B’s initial velocity is m s They are still separated by 85 m, but it has become easier to calculate the time interval in which Runner B has traveled the 85 m Known: vx m s; ax 0.10 m s2 ; x 85 m; for constant acceleration, t 10.s SOLVE First, we calculate how far Runner B travels during constant acceleration x vx 0t 21 ax t m s 10.s 21 0.10 m s2 10.s 5.0 m After the time interval of 10.s , Runner B is traveling vx vx ax t m s 0.10 m s 10.s 1.0 s Now Runner B has x 85 m 5.0 m 80.m left to run At vx 1.0m s, t x 80.m 80.s v x 1.0 m s Motion in One Dimension 2.51 Runner B’s total time to overtake Runner A is then ttotal ta tv 10.s 80.s 90.s REFLECT We would get exactly the same result if we considered the velocities of the two runners with respect to the track Notice that we have not taken into account the actual distance traveled with respect to the track This would become important if the question asked for the winner of the race! 100 ORGANIZE AND PLAN Since the rocks are in free fall, we can use v2 v02 2ay y to find the final speed of each rock Then we can find the definition of acceleration to find the time We’ll use subscripts and to refer to the two rocks Known: vy 01 10.m s ; vy 02 10.m s; y 12 m ; a y g 9.80 m s SOLVE vy2 vy20 2a y y For the first rock, v12 v01 ay y v1 v01 a y y 10.m s 9.80 m s 12 m 18.3 m s We chose the negative root because the rock is falling downward The elapsed time is y t1 v1 v01 t 12 m 2y 2.89 s v1 v01 18.3 m s 10.m s For the second rock, v1 10.m s 9.80 m s 12 m 18.3 m s But t2 12 m 18.3 m s 10.m s 0.85 s REFLECT Objects in free fall launched from the same point at the same speed (in the vertical direction) strike the ground with the same final velocity! However, they strike the ground at different times We’ll look at this in another way when we study kinetic and potential energy in Chapter 101 ORGANIZE AND PLAN Here the two trains must end up in the same place at the same time We will use the kinematic equations x vx t to find displacement at constant velocity and x vx t 21 a x t to find displacement at constant acceleration We’ll set these two equations equal to each other and solve for t Then we can find where the two trains meet again by substituting t into one of the equations for position We’ll indicate variables for the two trains by using subscripts and Known: vx1 11 m s; vx02 m s; ax 1.5 m s2 SOLVE For the first train, x1 vx1t t x1 v x1 For the second train, x v x 01t 21 a x t Since vx 01 m s , x 21 a x t 2 2.52 Chapter Setting the two equations equal, v x1t 21 a y t t 21 ax t vx1 t 0s when the second train initially passes the first Therefore, when the two trains meet again, a t x t vx1 2vx1 11 m s 14.7 s 15 s ax 1.5 m s2 Now we substitute t into one of the original equations: x vx1t 11 m s 14.7 s 160 m REFLECT At an acceleration of 1.5m s , the second train would have a velocity equal to that of the first train in a little over s, even though the second train would still be behind the first Since the second train is still accelerating, it is reasonable to think that it will overtake the first train in about 15 s 102 ORGANIZE AND PLAN In this problem, we establish an origin where the front ends of the trains meet We write equations for the position of each train, set the equations equal, and solve for t In (a) this is straightforward, using the formula for displacement with constant velocity In (b), however, we use the quadratic formula to solve directly for t We’ll use the subscripts and with variables that apply to train and train Known: (a) x0 m; vx1 22.5 m s; vx2 22.5 m s; x1 110.m; x2 110.m (b) x0 m s; vx 01 22.5 m s; ax1 1.0 m s ; vx2 vx02 22.5 m s; x1 110.m; x2 110.m SOLVE First we establish an origin We declare that the locomotives of each train pass each other at the origin at t s Then the position of locomotive is x1 x0 vx t Since locomotive is traveling the opposite direction, its position is x2 x0 vx t Now we write expressions for the locations of the ends of the trains: end1 x0 vx t 110.m and end2 x0 vx t 110.m Since the ends are in the same position at the same time, we can set these two equations equal to each other: x0 vx t 110.m x0 vx t 110.m Simplifying and collecting like terms, 2vx t 220.m and vx t 110.m t Since t 110.m 4.89 s 22.5 m s 110.m is just the time it takes one whole train to pass the origin, both ends will pass the origin at this vx elapsed time, but traveling in opposite directions (b) Now, if one train accelerates, the position of the first locomotive is end1 x0 vx 01t 21 a x t The equation for the position of locomotive at constant velocity remains as it was in (a), end2 x0 vx t 110.m Motion in One Dimension 2.53 Setting the equations for the positions of the two locomotives equal, we get x0 vx 01t 21 a x t x0 v x t 110.m Simplifying and putting in the general form of a quadratic, a x t 2v x t 220 m Using the quadratic formula, t 2v x 2vx 2 21 ax 220.m 21 ax 45 m s 2 1.0 2 22.5 m s m s2 220.m 1.0 m s2 The roots of this equation are t 4.65 s and t 94.6 s In this problem, only the positive root has any physical meaning, so t 4.65 s REFLECT The time for the ends of the trains to meet when one train is accelerating is less than when both trains are at constant velocity, which is what we would expect If we extend the conditions of (b) backward in time, we find that the other root of the quadratic refers to a point in time past when the first train passed the second train while traveling in the same direction as the second train at a high velocity, slowing down, reversing direction, and then meeting the second train again as this problem states 103 ORGANIZE AND PLAN Since we are only given two known variables in this free-fall problem, we must consider the first half of the motion of the ball, where it reaches maximum height and a final velocity of vy m s We’ll use v2y vy20 2ay y to find initial velocity Then we can use the definition of acceleration to find the time to reach maximum height From symmetry considerations, we can double the time to find when the ball is caught Known: y 17.2 m 1.6 m 15.6 m; vy m s; a y g 9.80 m s SOLVE First we find initial velocity: v2y vy20 2ay y vy 2a y y 2 9.80 m s2 15.6 m 17.49 m s Now, to find time to maximum height, t vy vy0 ay 0m s 17.49 m s 9.80 m s2 1.784 s Since the ball was caught at the same height at which it was thrown upward, we know from symmetry that the ball takes the same time interval to come back down, and ttotal 1.784 s 3.57 s With 4.8 s on the clock, this leaves 1.2 s for the opponent to take a shot REFLECT Realistically, if the opposing player were facing the basket and could make the catch and release a shot in one smooth motion, there might be a chance of a basket Otherwise the game is over ... though we divide the distance traveled by two in each successive step, we also divide the time it takes by two The speed therefore remains constant Therefore, if you cover half the distance to... we can calculate the average velocity for the entire run We’ll use subscript “1” for running forward, subscript “2” when jogging backward, and no subscript for a variable that applies to the entire... same time, we can set the equations for each animal’s position equal to each other and solve for t We’ll use subscript (1) for the cheetah and subscript (2) for the zebra Known: x10 m; x20
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