Solutions 3-1 (a) Distance hiked = b + c km b km (b) Displacement is a vector representing Paul’s change in position Drawing a diagram of Paul’s trip we can see that displacement –c km his displacement is b + (–c) km east = (b –c) km east (c) Distance = km + km = km; Displacement = (5 km – km) east = km east 3-2 (a) From v d t v x t x We want the answer in m/s so we’ll need to convert 30 km to meters and t to seconds: x 30,000 m m 30.0 km 1000 30, 000 m; 8.0 min 160mins 480 s Then v 63 m km s t 480 s Alternatively, we can the conversions within the equation: 1000 m x 30.0 km km v 63 m s t 8.0 min 160mins (b) v In mi/h: mi h 30.0 km 1.61 18.6 mi; 8.0 min 601min 0.133 h Then v km x 18.6 mi 140 mi h t 0.133 h mi x 30.0 km 60 x 30.0 km 1.61 km mi mi Or, v h 1.61 km 140 h Or, v 140 mi h h t t 8.0 8.0 min 60 There is usually more than one way to approach a problem and arrive at the correct answer! d L v t t L 24.0 m (b) v 40 ms t 0.60 s 3-3 (a) From v d x v t t x 0.30 m (b) v 30 ms t 0.010 s 3-4 (a) From v d 2 r t t 2 r 2 (400m) 63 ms (b) v t 40s 3-5 (a) v © Paul G Hewitt and Phillip R Wolf 3-1 d t 3-6 (a) t = ? From v (b) t t h v h 508 m 34 s v 15 ms (c) Yes At the beginning of the ride the elevator has to speed up from rest, and at the end of the ride the elevator has to slow down These slower portions of the ride produce an average speed lower than the peak speed 3-7 (a) t = ? Begin by getting consistent units Convert 100.0 yards to meters using the conversion factor on the inside cover of your textbook: 0.3048 m = 1.00 ft d d 91.4 m 3ft 0.3048 m t Then 100.0 yards 1yard 91.4 m From v 1ft t v v d 91.4 m (b) t 15 s v 6.0 ms d d L t t v c L 1.00 m 3.33 10-9s 3.33 ns (This is 13 billionths of a second!) (b) t v 3.00 10 ms 3-8 (a) t = ? From v d d vt t (b) First, we need a consistent set of units Since speed is in m/s let’s convert minutes to seconds: 3-9 (a) d = ? From v 60 s 5.0 1min 300 s Then d v t 7.5 ms 300s 2300 m 3-10 (a) v v0 vf v 2 (b) d ? From v (c) d d vt d vt t (1.5s) 1.5 m 2.0 ms vt 2 3-11 (a) d ? From v d t vt v v v d vt f t t 12 ms (8.0s) vt 48 m (b) d 2 3-12 (a) d ? From v d t vt v v v d vt f t t (b) First get consistent units: 100.0 km/h should be expressed in m/s (since the time is in 27.8 ms (8.0 s) vt km 1h 1000 m m 110 m seconds) 100.0 h 3600 s km 27.8 s Then, d 2 © Paul G Hewitt and Phillip R Wolf 3-2 v v2 v1 t t 15 km 25 km Since our time is in seconds we need to convert km to (b) v 40 km h h h h 3-13 (a) a m s : m v 6.94 s 0.35 m2 s t 20 s Alternatively, we can express the speeds in m/s first and then the calculation: 11.1 ms 4.17 ms 1hr 1000 m m km 1hr 1000 m m 15 km 4.17 and 40 11.1 Then a 0.35 m2 hr 3600 s km s hr 3600 s km s s 20s 1hr m 25 km 3600 1000 6.94 ms Then a h s km v v2 v1 t t (b) To make the speed units consistent with the time unit we’ll need v in m/s: 3-14 (a) a 1hr m v v2 v1 20.0 km 5.0 km 15.0 km 3600 1000 4.17 ms Then a h h h s km m v2 v1 4.17 s 0.417 m2 s t 10.0 s An alternative is to convert the speeds to m/s first: 1hr m 1hr m v1 5.0 km 3600 1000 1.4 ms ; v2 20.0 km 3600 1000 5.56 ms h s km h s km Then a (c) d vt 5.56 ms 1.4 ms v2 v1 0.42 m2 s t 10.0 s 1.4 ms 5.56 ms v1 v2 t 10.0 s 35 m Or, 2 d v1t 12 at 1.4 ms (10.0 s) 12 0.42 m2 (10.0 s)2 35 m s v vf v0 v v t t t t m v 26 s (b) a 1.3 m2 s t 20s 3-15 (a) a (c) d ? From v d t 26 ms ms v v d vt f t 20 s 260 m Or, d v0t 12 at 26 ms (20 s) 12 1.3 m2 (20 s)2 260 m s (d) d = ? Lonnie travels at a constant speed of 26 m/s before applying the brakes, so d vt 26 ms (1.5 s) 39 m v vf v0 v v t t t t m v 72 s 6.0 m2 (b) a s t 12 s 3-16 (a) a © Paul G Hewitt and Phillip R Wolf 3-3 (c) d ? From v 72 ms ms v v d vt f t (12 s) 430 m d t Or, d v0t 12 at 72 ms (12 s) 12 6.0 m2 (12 s)2 430 m 3-17 (a) t ? From v (b) t d t t s d L 2L v v f v v 2L 2(1.4 m) 0.19 s v 15.0 ms v v v 3-18 (a) v f 350 ms (b) v 175 ms Note that the length of the barrel isn’t needed—yet! (c) From v d t t d L 0.40 m 0.0023 s 2.3 ms v v 175 ms 3-19 (a) From v d t v v v v d vt f t = t 25 ms 11 ms v v t = (b) d (7.8 s) 140 m 3-20 (a) v = ? There’s a time t between frames of s, 24 so v= d t 24 x (That’s 24x per s s x 24 second.) (b) v 24 1s x 24 1s (0.15 m) 3.6 ms 3-21 (a) a = ? Since time is not a part of the problem we can use the formula vf v0 2ad and v2 solve for acceleration a Then, with v0= and d = x, a 2x 1.8 10 ms v2 (b) a 2x 2(0.10 m) 1.6 10 15 m s2 1.8 10 ms ms vf v0 1.1 10-8 s 11 ns (c) t ? From vf v0 at t a 1.6 1015 m2 s Or, from v d t © Paul G Hewitt and Phillip R Wolf d L 2L 2(0.10 m) -8 t v v 1.1 10 s f v (v 0) 1.8 10 m s 3-4 d v0 vf 2d t with v0 vf t t 3-22 (a) vf =? From v (b) af =? From d v0 t 12 at with v0 d 12 at (c) vf a 2d t2 2d 2(402 m) 2d 2(402 m) 181 ms ; a 40.6 m2 s t 4.45 s t (4.45 s) 3-23 (a) d=? From v v v v V d vt f t = t d t 110 ms 250 ms v V t = (b) d (3.5 s) 630 m 3-24 (a) t ? Let's choose upward to be the positive direction v v 0v v From vf v0 at with vf and a g t f a g g (b) t m v 32 s 3.3 s g 9.8 m2 s m 2 32 s d v v v 0 v v (c) d ? From v d vt f t 52 m g 2g 9.8 m t 9.8 (3.3 s) s We get the same result with d v0t at 2 (3.3 s) 32 ms 3-25 (a) v0 = ? When the potato hits the ground y = From d v0t 12 at (b) v0 12 gt y v0t 12 gt 9.8 (12 s) 59 m s2 t v0 12 gt m In s 2 m s2 52 m v0 12 gt km s mi 3600 130 mi mi/h, 59 ms 1000 m 1.61 km 1h h 3-26 (a) t = ? Choose downward to be the positive direction From From d v0 t 12 at with v0 0, a g and d h h 12 gt (b) t t 2h g 2h 2(25m) 2.26 s 2.3 s g 9.8 m2 s (c) vf vo at gt 9.8 m2 (2.26 s) 22 m s s 2 m m Or, from 2ad vf v0 with a g, d h, and v0 vf 2gh 9.8 s2 (25 m) 22 s © Paul G Hewitt and Phillip R Wolf 3-5 3-27 (a) v0 = ? Let’s call upward the positive direction Since the trajectory is symmetric, vf = –v0 gt Then from vf v0 at, with a g v0 v0 gt 2v0 gt v0 9.8 m2 (4.0s) gt s (b) v0 20 ms 2 20 ms d v v v d vt f t t (2.0 s) 20 m 2 t We use t = 2.0 s because we are only considering the time to the highest point rather than the whole trip up and down (c) d ? From v 3-28 (a) v0 = ? Let’s call upward the positive direction Since no time is given, use vf v0 2ad with a = –g, vf = at the top, and d = (y – m) v02 2(g)(y 2m) v0 2g(y m) (b) v0 2g(y m ) 9.8 m2 (20 m m) 18.8 ms 19 ms s 3-29 (a) Taking upward to be the positive direction, from 2ad vf2 v02 with a g and d h vf v02 2gh So on the way up vf v02 2gh (b) From above, on the way down vf v02 2gh, same magnitude but opposite direction as (a) (c) From a vf v0 t t 2 vf v0 v0 2gh v0 v0 v0 2gh a g g (d) vf v02 2gh 16 ms 9.8 m2 (8.5 m) 9.5 ms t s m m vf v0 9.5 s 16 s 2.6 s a 9.8 m2 s 3-30 (a) vf = ? Taking upward to be the positive direction, from 2ad vf2 v02 with a g and d h vf v02 2gh The displacement d is negative because upward direction was taken to be positive, and the water balloon ends up below the initial position The final velocity is negative because the water balloon is heading downward (in the negative direction) when it lands 2 vf v0 vf v0 v0 2gh v0 v0 v0 2gh (b) t = ? From a t t a g g (c) vf = ? Still taking upward to be the positive direction, from 2ad vf2 v02 with initial velocity = –v0 , a g and d h vf2 v02 2gh vf v02 2gh We take the negative square root because the balloon is going downward Note that the final velocity is the same whether the balloon is thrown straight up or straight down with initial speed v0 © Paul G Hewitt and Phillip R Wolf 3-6 5.0 ms 9.8 sm (11.8 m) 16 ms for the balloon whether it is (d) vf v02 2gh tossed upward or downward For the balloon tossed upward, 16 ms ms v v t f 2.1 s a 9.8 m2 s 3-31 (a) Call downward the positive direction, origin at the top From d v0 t 12 at with a g, d ² y h h v0 t 12 gt From the general form of the quadratic formula x g , a b v0 , and c h, which gives t 12 gt v0 t h b b 4ac we identify 2a v0 v0 (h) v g g To get a positive value for the time we take the positive root, and get t v0 + v0 + 2gh v0 2gh g g (b) From 2ad vf2 v02 with initial velocity v0 , a g and d h vf2 v02 2gh vf v02 2gh v v 2gh v0 + 2gh Or you could start with vf v0 at v0 g g (c) t v0 v0 2gh g t2 9.8 sm2 0.58 s ; 2 3-32 (a) From d v0 t 12 at 2(d v0t) 3.2 ms 2 9.8 sm (3.5 m) 3.2 ms 9.8 sm (3.5 m) 8.9 ms vf v02 2gh (b) a 3.2 ms a 2(d v0 t) t2 120 m 13 ms ·5.0 s (5.0 s)2 4.4 m s2 (c) vf v0 at 13 ms 4.4 sm2 (5s) 35 ms km s mi 3600 78 mi This is probably not a safe speed for driving in (d) 35 ms 1000 m 1.61 km 1h h an environment that would have a traffic light! v v 2x v0 3-33 (a) From x vt f t vf t x x vf v0 (2 t v0 ) v0 t 2v0 x v a 2 (b) t t t t t © Paul G Hewitt and Phillip R Wolf 3-7 2x 2(95m) v0 13 ms 3.0 ms t 11.9s m m 95m 13 ms vf v0 3.0 s 13 s x v m a 2 2 0.84 or a 0.84 m2 s s t t 11.9s t 11.9s (11.9s) (c) vf 3-34 (a) From 2ad vf2 v02 with d L vf v02 2aL This is Rita’s speed at the bottom of the hill To get her time to cross the highway: From v d (b) t v02 2aL 25m 2 3.0 ms 1.5 sm2 (85m) d t t d d vf v02 2aL 1.54s 3-35 (a) Since v0 is upward, call upward the positive direction and put the origin at the ground Then From d v0 t 12 at with a g, d ² y h h v0 t 12 gt 12 gt v0 t h From the general form of the quadratic formula x a g , b v0 , and c h, which gives t b b 4ac we identify 2a v0 v0 g (h) v g v0 2gh g (b) From 2ad vf2 v02 with a g and d h vf2 v02 2gh vf v02 2gh (c) t v0 v0 2gh g 22 ms 22 ms 2 9.8 sm (14.7m) 9.8 sm2 0.82 s or 3.67 s So Anthony has to have the ball leave his had either 0.82s or 3.67s before midnight The first time corresponds to the rock hitting the bell on the rock’s way up, and the second time is for the rock hitting the bell on the way down vf v02 2gh 22 ms 9.8 sm (14.7m) 14 ms 2 3-31 (a) v1 = ? The rocket starts at rest and after time t1 it has velocity v1 and has risen to a height h1 Taking upward to be the positive direction, from vf v0 at with v0 v1 at1 (b) h1 = ? From d v0 t 12 at with h1 d and v0 h1 12 at12 (c) h2 = ? For this stage of the problem the rocket has initial velocity v1, vf = 0, a = –g and the distance risen d = h2 vf2 v02 v12 v12 (at1 )2 a t12 h2 2a 2(g) 2g 2g 2g (d) tadditional = ? To get the additional rise time of the rocket: From v v v v v1 at1 a f t additional f t a g g (e) The maximum height of the rocket is the sum of the answers from (a) and (b) = a2t hmax h1 h2 12 at12 12 at12 ag 2g From 2ad vf2 v02 d © Paul G Hewitt and Phillip R Wolf 3-8 (f) tfalling = ? Keeping upward as the positive direction, now v0 = 0, a = –g and d = –hmax From d v0 t 12 at hmax 12 (g)t 12 at12 ag g 2hmax g t falling (g) t total t1 t additional t falling t1 at12 (g a) g a(g + a) at1 t a(g + a) g g 120 (1.70 s) a t h 2g 9.8 (h) vruns out of fuel v1 at1 120 m2 (1.70 s) 204 ms ; h1 12 at12 s hadditional m s2 2 2 t additional t1 g 120 (1.70 s) 173 m m s2 2 2123 m m s2 m at1 120 s2 (1.70 s) 20.8 s g 9.8 m2 s hmax 173 m + 2123 m 2296 m 2300 m t falling 2hmax 2(2300 m) 21.7 s g 9.8 m2 s t total t1 tadditional tfalling 1.7 s 20.8 s 21.7 s 44.2 s total distance xx 2x 1.14 x t total time t 0.75t 1.75t 140 km (b) v 1.14 x 1.14 80 km hr t hr 3-32 v total distance d From v d vt total time t dwalk d jog vwalk t walk v jog t jog v(30 min) 2v(30 min) 3v(30 min) So v 1.5 v t walk t jog t walk t jog 30 min 30 2(30 min) 3-33 (a) v (b) v 1.5v 1.5 1.0 ms 1.5 ms s 5400 m = 5.4 km (c) dto cabin vt total v (t walk t jog ) 1.5 ms (30 +30 min) 160 total distance d From v d vt total time t d dfast vslowtslow vfast t fast v(1 h) 4v(1 h) 5v(1 h) So v slow 2.5 v tslow t fast tslow t fast h 1 h 2h 3-34 (a) v (b) v 2.5v 2.5 25 km 63 km h h © Paul G Hewitt and Phillip R Wolf 3-9 total distance d d x From v t total time t v v d1 d2 2x 2x So v v v t1 t x x x v1 v1 v1v2 v1 v2 3-35 (a) v v1v2 1.5v v(1.5v) 2.5v 1.2v Note that the average velocity is biased toward 1.5v v v2 v1 the lower speed since you spend more time driving at the lower speed than the higher speed 2 (b) v 1.2v 1.2 28 km 34 km h h 3-36 (a) dAtti =? From VAtti dAtti t dAtti Vt The time that Atti runs = the time that Judy x x V walks, which is t So dAtti V x v v v 4.5 ms V (150 m) 450 m (b) X x m v 1.5 s 3-37 v d 3m ms t 1.5 s 3-38 h = ? Call upward the positive direction From vf v0 2ad with d h, vf and a g 2 14.7 ms v v0 v0 v0 h f 11 m 2a 2(g) 2g 9.8 m s d 3-39 d ? From v t 27.5 ms v0 vf d vt t (8.0 s) 110 m 3-40 t ? Let's take down as the positive direction From d v0 t 12 at with v0 and a g d 12 gt t 2d 2(16 m) 1.8 s g 9.8 m2 s 3-41 a m m v vf v0 12 s s m2 s t t 3s 3-42 a m m v vf v0 75 s s 30 m2 s t t 2.5 s 3-43 d = ? With v0 0, d v0t 12 at becomes d 12 at © Paul G Hewitt and Phillip R Wolf 3-10 2.0 (8.0 s) 64 m m s2 3-44 a ? With v0 0, d v0 t 12 at becomes d 12 at a 3-45 d = ? With v0 0, d v0t 12 at becomes d 12 at 2d t 2(5.0 m) (2.0 s)2 2.5 m2 s 3.5 (5.5 s) 53 m m s2 3-46 v0 = ? Here we’ll take upwards to be the positive direction, with a =g and vf = From vf v0 2ad v0 vf 2(g)d v0 2gd 9.8 m2 (3.0 m) 7.7 ms s 3-47 t=? We can calculate the time for the ball to reach its maximum height (where the velocity will be zero) and multiply by two to get its total time in the air Here we’ll take upward to be the positive direction, with a =g m vf v0 vf v0 v0 v0 18 s From a t 1.84 s This is the time to reach t a g g 9.8 m2 s the maximum height The total trip will take 1.84 s = 3.7 s, which is less than s Alternatively, this can be done in one step with by recognizing that since the trajectory is symmetric vf = –v0 Then from vf v0 at, with a g v0 v0 gt 2v0 gt m 2v0 18 s t 3.7 s g 9.8 m2 s 3-48 v0 = ? Since she throws and catches the ball at the same height, vf v0 Calling upward the positive direction, a = –g 9.8 m2 (3.0 s) gt s From vf v0 at v0 v0 (g)t 2v0 gt v0 15 ms 2 3-49 For a ball dropped with v0 = and a = +g (taking downward to be the positive direction), dfallen, 1st v0t 12 at second 9.8 (1 s) 4.9 m At the beginning of the 2 m s2 we have v0 = 9.8 m/s so dfallen, 2nd ratio second nd second v0t 12 at 9.8 ms (1 s) 12 9.8 m2 (1 s)2 14.7 m The s dfallen, 2nd second dfallen, 1st second 14.7 m More generally, the distance fallen from rest in a time 4.9 m t is d gt in the next time interval t the distance fallen is 2 dfrom time t to 2t v0 t 12 at (gt)t 12 gt 23 gt The ratios of these two distances is dfrom time t to 2t dfrom rest in time t © Paul G Hewitt and Phillip R Wolf gt 2 gt 2 3-11 3-50 h ? Call upward the positive direction From vf v0 2ad with d h, vf and a g 1,000 ms vf v0 v0 v0 h 2a 2(g) 2g 9.8 m2 2 51, 000 m > 50 km s 3-51 h ? With d h, v0 22 m , s a –g and t 3.5 s, d v0 t 12 at becomes h (22 ms )(3.5 s) 12 9.8 m2 (3.5 s)2 17 m 3-52 t =? From v d t 3-53 t ? From a t d 65 m 5.0 s v 13 ms v vf v0 t t 3-54 (a) t ? From v d t s t t m m vf v0 28 s s 4.0 s a 7.0 m2 s d d 2d v v f v v (b) a = ? With v0 and vf v, vf – v0 2ad becomes a 2 v2 2d 28 ms 2d 2(140 m) v2 10 s; a 2.8 m2 (c) t s v 2d 2(140 m) 28 ms ms 25 ms d v0 vf t (5.0 s) 63 m 3-55 d ? From v d vt t km d d 2462 mi 0.621 mi 3-56 t = ? From v t 8.5 t v 28,000 km h 60 h 3-57 a ? With vf 0, vf – v0 2ad becomes mi km 1000 m 1h v0 220 h 0.621 mi km 3600 s a 2d 2(800 m) 3-58 From d v0 t 12 at x t 2 –6.05 m –6 m s2 s2 12 at v0 t d From the general form of the quadratic formula b b 4ac we identify a 12 a, b v0 , and c h, which gives 2a v0 v0 a 12 a(d) v0 the positive root, which gives us t © Paul G Hewitt and Phillip R Wolf v0 2ad a 45 ms To get a positive answer for t we take 45 ms 2 3.2 sm (440m) 3.2 sm2 3-12 7.7 s 3-59 v0 = ? The candy bar just clears the top of the balcony with height 4.2m + 1.1m = 5.3 m With vf 0, vf – v0 2ad with v0 and ² y positive and a g v0 vf 2(g)h v0 2gh 9.8 m2 (5.3m) 10.19 ms 10.2 ms The total time is the time for the s way to the top of the balcony rail plus the time to fall 1.1 m to the floor of the balcony 2d 2(5.3 m) t up ? From d vf t 12 at with vf and a g d 12 (g)t t up 1.04 s g 9.8 sm2 t down ? From d v0t 12 at with v0 0, a g and d ² y h h 12 (g)t t down 2h 2(1.1 m) 0 g 9.8 sm2 So t total tup tdown 1.040 s 0.47 s 1.51s An alternative route is: Since v0 is upward, call upward the positive direction and put the origin at the ground Then From d v0 t 12 at with a g, d ² y 4.2m d v0 t 12 gt 12 gt v0 t d From the general form of the quadratic formula x a g , b v0 , and c d, which gives t 10.19 ms 10.19 ms 2 9.8 sm (4.2m) 9.8 sm2 b b 4ac we identify 2a v0 v0 g (d) v g v0 2gd g 0.57 s or 1.51s The first answer corresponds to the candy reaching 4.2 m but not having gone over the top balcony rail yet The second answer is the one we want, where the candy has topped the rail and arrives 4.2 m above the ground 3-58 Consider the subway trip as having three parts—a speeding up part, a constant speed part, and a slowing down part dtotal dspeeding up dconstant speed dslowing down For dspeeding up , v0 0, a 1.5 m2 and t 12 s, so d v0t 12 at s For dconstant speed vt From the speeding up part we had v0 0, 1.5 (12 s) 108 m m s2 a 1.5 m2 and s so v v0 at 1.5 m2 (12 s) 18 ms and so d 18 ms (38 s) 684 m s t 12 s For dslowing down , vf 0, a –1.5 m2 and t 12 s, so d vf t 12 at 12 -1.5 m2 (12 s)2 108 m s s So dtotal dspeeding up dconstant speed dslowing down 108 m 684 m 108 m 900 m 3-59 One way to approach this is to use Phil’s average speed to find how far he has run during the time it takes for Mala to finish the race 100.0 m d From v dPhil vPhilt Mala (12.8 s) 94.1 m Since Phil has only t 13.6 s traveled 94.1 m when Mala crosses the finish line, he is behind by 100 m 94.1 m 5.9 m m © Paul G Hewitt and Phillip R Wolf 3-13 3-60 t = ? The time for Terrence to land from his maximum height is the same as the time it takes for him to rise to his maximum height Let’s consider the time for him to land from a height of 0.6 m Taking down as the positive direction: From d v0 t 12 at with v0 and a g d 12 gt t 2d 2(0.6 m) 0.35 s g 9.8 m2 s His total time in the air would be twice this amount, 0.7 s 3-61 v d mi 80 mi h 1h t 45 s 3600 s total distance d d If we call the distance she drives d, then from v t total time t v dthere dback 2d 2d v v So v v v there back d d back there t there t back vback vthere d 3-62 v vthere vthere vback vback v there vback 2400 km 40 km 60 km h 48 km Note that the average velocity is biased h h 2 2 h km km 60 h 40 h 100 kmh toward the lower speed since Norma spends more time driving at the lower speed than at the higher speed © Paul G Hewitt and Phillip R Wolf 3-14 ... 15 ms (c) Yes At the beginning of the ride the elevator has to speed up from rest, and at the end of the ride the elevator has to slow down These slower portions of the ride produce an average... To get the additional rise time of the rocket: From v v v v v1 at1 a f t additional f t a g g (e) The maximum height of the rocket is the sum of the answers from (a) and (b) =... between frames of s, 24 so v= d t 24 x (That’s 24x per s s x 24 second.) (b) v 24 1s x 24 1s (0.15 m) 3.6 ms 3-21 (a) a = ? Since time is not a part of the problem we