Conceptual Physics 12th Edition Hewitt Solutions Manual PDF

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Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual 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...

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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 = 5 km + 2 km = 7 km; Displacement = (5 km – 2 km) east = 3 km east. 3-2. (a) From v  (b) 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 8 min 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 . 1 km s t 480 s Alternatively, we can do the conversions within the equation: 1000 m x 30.0 km  1 km v   63 m . s t 8.0 min 160mins

In mi/h:

1 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

1 mi x 30.0 km 60 min x 30.0 km  1.61 km 1 mi mi Or, v    1 h  1.61 km  140 h . Or, v    140 mi . h 1 h t t 8.0 min 8.0 min 60 min

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

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3-6. (a) t = ? From v  (b) t 

d t

t 

h 508 m   34 s. v 15 ms

h . v

(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 Then 100.0 yards  1yard   91.4 m. From v  t   . 1ft t v v d 91.4 m (b) t    15 s. v 6.0 ms





3-8. (a) t = ? From v 

d d L t   . t v c L 1.00 m  3.33  10-9s  3.33 ns. (This is 3 13 billionths of a second!) (b) t   v 3.00  10 8 ms

3-9. (a) d = ? 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: 5.0 min  1min  300 s. Then d  v t  7.5 ms  300s  2300 m. 60 s

3-10. (a) v 

v0  vf v  . 2 2

(b) d  ? From v  (c) d 



d vt  d  vt  . t 2 (1.5s)  1.5 m.



2.0 ms vt  2 2

3-11. (a) d  ? From v 

 

d t

vt v v   0  v  d  vt   0 f  t   t .   2   2  2

12 ms (8.0s) vt  48 m. (b) d   2 2

3-12. (a) d  ? From v 

d t

vt v v   0  v  d  vt   0 f  t   t .  2   2  2











(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  1 km  27.8 s . Then, d   2 2 © Paul G. Hewitt and Phillip R. Wolf

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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 20 s t Alternatively, we can express the speeds in m/s first and then do 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 1 km s hr 3600 s 1 km s s 20s 1hr m 25 km  3600  1000  6.94 ms . Then a  h s 1 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 1 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 1 km h s 1 km

Then a  (c) d  vt 





5.56 ms  1.4 ms v2  v1   0.42 m2 . s 10.0 s t

 1.4 ms  5.56 ms  v1  v2 t  10.0 s  35 m. Or, 2 2  









d  v1t  12 at 2  1.4 ms (10.0 s)  12 0.42 m2 (10.0 s)2  35 m. s

v vf  v0 0  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  0 ms  v v   d  vt   0 f  t    20 s  260 m.  2  2  







Or, d  v0t  12 at 2  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 0  v v    . t t t t m v 72 s   6.0 m2 . (b) a  s 12 s t

3-16. (a) a 

© Paul G. Hewitt and Phillip R. Wolf

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(c) d  ? From v 

 72 ms  0 ms  v v   d  vt   0 f  t    (12 s)  430 m.  2  2  



d t



Or, d  v0t  12 at 2  72 ms (12 s)  12 6.0 m2 (12 s)2  430 m.

3-17. (a) t  ? From v  (b) t 

t 

d t

2L 2(1.4 m)   0.19 s. v 15.0 ms

s

 

d L 2L  v v  . f 0 v v 2

v v  v 3-18. (a) v   0 f   .  2  2 350 ms (b) v   175 ms . Note that the length of the barrel isn’t needed—yet!

2

(c) From v  3-19. (a) From v 

d t

d t

 t

d L 0.40 m    0.0023 s  2.3 ms. v v 175 ms

v v   v  v  d  vt   0 f  t =  0 t.  2   2 

 25 ms  11 ms   v  v t = (b) d   0  (7.8 s)  140 m.   2  2  

3-20. (a) v = ? There’s a time t between frames of

   

1 s, 24

so v=

d  t

second.)

 24 1 x. (That’s 24x per  s  s  x

1 24

(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 2  v0 2  2ad and

v2 solve for acceleration a. Then, with v0= 0 and d = x, a  . 2x



1.8  10 7 ms v2 (b) a   2x 2(0.10 m)

  1.6  10 2

15 m s2

.





1.8  10 7 ms  0 ms vf  v0   1.1  10-8 s  11 ns. (c) t  ? From vf  v0  at  t  a 1.6  1015 m2   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.  0 f v (v  0) 1.8  10 7 m s 2 

 



s

3-4

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

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d  v0  vf  2d  t with v0  0  vf  .  t  2  t

3-22. (a) vf =? From v 

(b) af =? From d  v0 t  12 at 2 with v0  0  d  12 at 2 (c) vf 

a

2d 2(402 m) 2d 2(402 m)   181 ms ; a  2   40.6 m2 . 2 s t 4.45 s t (4.45 s)

3-23. (a) d=? From v 

2d t2

.

v v   v V  d  vt   0 f  t =  t.  2   2 

d t

 110 ms  250 ms   v V  t = (b) d    (3.5 s)  630 m.   2  2  

3-24. (a) t  ? Let's choose upward to be the positive direction. v v 0v v From vf  v0  at with vf  0 and a  g  t  f 0   . a g g (b) t 

m v 32 s   3.3 s. g 9.8 m2



s



  9.8 (3.3 s)

2 32 s d v v   v  0  v v (c) d  ? From v   d  vt   0 f  t      52 m.   2   2   g  2g 2 9.8 m t 2

We get the same result with d  v0t  at  1 2

2

 (3.3 s)  32 ms



3-25. (a) v0 = ? When the potato hits the ground y = 0. From d  v0t  12 at 2

(b) v0  12 gt 

1 2

9.8 (12 s)  59

 y  v0t  12 gt 2 m s2

 0  t v0  12 gt

m . In s



s

1 2

2

m s2

 52 m.

 v0  12 gt.

1 km s  1 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 2 with v0  0, a  g and d  h  h  12 gt 2

(b) t 

m 2

t 

2h . g

2h 2(25m)   2.26 s  2.3 s . g 9.8 m2 s

 

. (c) vf  vo  at  0  gt  9.8 m2 (2.26 s)  22 m s

 

 2 2 m m   Or, from 2ad  vf  v0 with a  g, d  h, and v0  0  vf  2gh  2 9.8 s2 (25 m)  22 s . s

© Paul G. Hewitt and Phillip R. Wolf

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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  . 2

 

9.8 m2 (4.0s) gt s (b) v0    20 ms . 2 2





20 ms d v v  v   d  vt   0 f  t   0  t  (2.0 s)  20 m.  2   2 t 2 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 2  v0 2  2ad with a = –g, vf = 0 at the top, and d = (y – 2 m).

 

 v02  2(g)(y  2m)  v0  2g(y  2 m).

(b) v0  2g(y  2 m )  2 9.8 m2 (20 m  2 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 2 vf  v0  v0  2gh  v0 v0  v0  2gh   . a g g

 

(d) vf   v02  2gh   16 ms

2

 

 2 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 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.

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5.0 ms   2 9.8 sm (11.8 m)  16 ms for the balloon whether it is

(d) vf   v02  2gh  

2

2

tossed upward or downward. For the balloon tossed upward, 16 ms  5 ms v v t f 0   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 2 with a  g, d  ² y  h  h  v0 t  12 gt 2 From the general form of the quadratic formula x  a

g , 2

b  v0 , and c  h, which gives t 

 12 gt 2  v0 t  h  0.

b  b 2  4ac we identify 2a

v0  v0 2  4

 (h)  v g 2

0

g To get a positive value for the time we take the positive root, and get

t

v0 + v0 2 + 2gh

 v0 2  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 2  2gh  0   v0 2 + 2gh . Or you could start with vf  v0  at  v0  g  0   g

(c) t 

v0  v0 2  2gh



t2



2

9.8 sm2

 0.58 s. ;

2

2

3-32. (a) From d  v0 t  12 at 2

2(d  v0t)

3.2 ms 2  2 9.8 sm (3.5 m)

3.2 ms   2 9.8 sm (3.5 m)  8.9 ms

g

vf   v02  2gh 

(b) a 

3.2 ms 



a

2(d  v0 t)

 4.4 m .

t2 2 120 m  13 ms ·5.0 s





(5.0 s)2

(c) vf  v0  at  13 ms  4.4 sm2 (5s)  35 ms .

.

s2

1 km s  1 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  0 f t  vf  2 t x x vf  v0 (2 t  v0 )  v0 2 t  2v0  x v  a     2 2  0  . (b) t t t t t 

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(c) vf 

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   2  or a    0.84 m2 .  0.84 2  2 s s t t  11.9s t 11.9s   (11.9s)

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 

(b) t 

d

v02  2aL



  2



25m

2 3.0 ms

1.5 sm2

(85m)

t 

d t

 1.54s.

d d  . vf v02  2aL

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 2 with a  g, d  ² y  h  h  v0 t  12 gt 2  12 gt 2  v0 t  h  0. From the general form of the quadratic formula x  a

g , 2

b  v0 , and c  h, which gives t 

b  b 2  4ac we identify 2a

v0  v0 2  4 g

 (h)  v g 2

0

 v0 2  2gh g

.

(b) From 2ad  vf2  v02 with a  g and d  h  vf2  v02  2gh  vf   v02  2gh. (c) t 

v0  v0 2  2gh g



22 ms 

22 ms 2  2 9.8 sm (14.7m) 2

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   2 9.8 sm (14.7m)  14 ms . 2

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  0  v1  at1 . (b) h1 = ? From d  v0 t  12 at 2 with h1  d and v0  0

 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 0  v12 v12 (at1 )2 a 2 t12  h2     . 2a 2(g) 2g 2g 2g (d) tadditional = ? To get the additional rise time of the rocket: From v v v v 0  v1 at1 a  f 0  t additional  f 0   . t a g g (e) The maximum height of the rocket is the sum of the answers from (a) and (b) = a2t 2 hmax  h1  h2  12 at12  1  12 at12 1  ag . 2g From 2ad  vf2  v02

d 

 

© Paul G. Hewitt and Phillip R. Wolf

3-8

Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual

Full file at https://testbankuniv.eu/Conceptual-Physics-12th-Edition-Hewitt-Solutions-Manual

(f) tfalling = ? Keeping upward as the positive direction, now v0 = 0, a = –g and d = –hmax. From d  v0 t  12 at 2

 t falling 

 hmax  12 (g)t 2

 

2  12 at12 1  ag    g

2hmax  g

(g) t total  t1  t additional  t falling  t1 

at12 (g  a) g

2

 a(g + a)

t at1  a(g + a) 1 . g g

  120  (1.70 s) a t h   2g 2 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 1

2

2

t additional 

2

m s2

1 2

t1 g

120 (1.70 s)  173 m. m s2

2

 2123 m.

m at1 120 s2 (1.70 s)   20.8 s. g 9.8 m2

hmax  173 m + 2123 m  2296 m  2300 m. s

t falling 

2hmax 2(2300 m)   21.7 s. g 9.8 m2

t total  t1  tadditional  tfalling  1.7 s  20.8 s  21.7 s  44.2 s. s

total distance xx 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  2 hr 

3-32. v 

3-33. (a) 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. 30 min 30 min 2(30 min) t walk  t jog t walk  t jog



