CSULA- PHYSICS 211

Pages to are hidden for

"CSULA- PHYSICS 211"

```					CSULA- PHYSICS 211                             Summer’12                  Solution: Material to Study                                      Chapter 2

OBJECTIVE QUESTIONS

OQ-2 (7e- None)
The arrow will reach its maximum height ~ 1.5s, then from that point where vyi = 0 to a point where vyf is 8 m/s, the arrow will take:
t = vyf /g ~ 0.8s. Therefore the total time will be ~2.3s, so the answer is (d).

OQ-13 (7e-Q11)
They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity of
magnitude vi . This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds
will also be the same.

OQ-16 (7e-Q14)
(b) Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point. So your ball must
travel a smaller distance to the passing point that the ball your friend throws.

CONCEPTUAL QUESTIONS
CQ-6 (7e-None)
(a). vyf = 0 m/s (b). ay = g = –9.8m/s2                                    (c). vyi = – vyf = – 5 m/s                                      (d). ay = g = –9.8m/s2

CQ-7 (7e-Q3)
No. Constant acceleration only. Yes. Zero is a constant.

CQ-9 (7e-Q5)
No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the
driver of B might have tramped hard on the gas pedal in the recent past to give car B greater acceleration just then.

PROBLEMS

P3 (7e-P3)
(a) Let d represent the distance between A and B. Let t1 be the time for which the walker has the higher speed in                                              d    . Let t 2
5.00 m s 
t1
d
represent the longer time for the return trip in 3.00 m s   d . Then the times are t1                         d       and t 2                   . The
t2                                   5.00 m s                   3.00 m s
average speed is:
Total distance                d d                            2d
vavg                                                   
Total time      d/  5.00 m s  d/  3.00 m s  8.00 m s d/  15.0 m 2 s2 
2  15.0 m 2 s2 
vavg                         3.75 m s
8.00 m s

(b) She starts and finishes at the same point A. With total displacement = 0, average velocity  0 .

P6 (7e-P6)
(a) At any time, t, the position is given by x   3.00 m s2  t 2 . Thus, at t i  3.00 s : xi   3.00 m s2   3.00 s2  27.0 m .

(b) At t f  3.00 s  t : x f   3.00 m s2   3.00 s  t  , or
2
x f  27.0 m   18.0 m s t   3.00 m s2   t        .
2

 x f  xi 
(c) The instantaneous velocity at t  3.00 s is:             v  lim                                                 
  lim 18.0 m s   3.00 m s  t  18.0 m s .
2
t  0
 t  t 0
P11 (7e-P9)
x f  xi       1 000 m  800 m
(a) Once it resumes the race, the hare will run for a time of             t                                  25 s .
vx                8ms

In this time, the tortoise can crawl a distance        x f  xi   0.2 m s  25 s  5.00 m .

(b) Hare was stationary for:       1000m
 25s  4975s  4.975 103 s
0.20m / s

P16 (7e-P11)
(a) Acceleration is constant over the first ten seconds, so at the end of this interval: v f  vi  at  0   2.00 m s2  10.0 s  20.0 m s .

Then a 0 so v is constant from t  10.0 s to t  15.0 s . And over the last five seconds the velocity changes to:

v f  vi  at  20.0 m s   3.00 m s2   5.00 s  5.00 m s .

at  0  0   2.00 m s2  10.0 s  100 m .
1 2           1
x f  xi  vi t 
2
(b) In the first ten seconds:
2             2
Over the next five seconds the position changes to: x f  xi  vi t  1 at 2  100 m   20.0 m s  5.00 s  0  200 m .
2
x f  xi  vi t  at  200 m   20.0 m s  5.00 s   3.00 m s2   5.00 s  262 m .
1 2                                   1
And at t  20.0 s :                                                                                    2

2                                     2

P17 (7e-P13)
dx
x  2.00  3.00t  t 2 , so v      3.00  2.00t , and a  dv  2.00
dt                            dt
At t  3.00 s :     (a)         x   2.00  9.00  9.00 m  2.00 m            (b) v   3.00  6.00 m s  3.00 m s       (c)   a 2.00 m s2

P20 (7e-P18)
(a)

(b)

(c)

(d)

(e)

(f) One way of phrasing the answer: The spacing of the successive positions would change with less regularity.
Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within
one drawing, the accelerations vectors would vary in magnitude and direction.
P29 (7e-P25)
     vxf  vxi  ax t                    vxf  vxi   5.60 m s2   4.20 s 
In the simultaneous equations: 

 we have



.

1

 x f  xi  vxi  vxf
           2
   t

1
 62.4 m  vxi  vxf  4.20 s 
             2
                   

So substituting for vxi gives: 62.4 m   vxf   5.60 m s2   4.20 s  vxf   4.20 s

1
2                                                        →        14.9 m s  vxf 
1
2
 5.60 m s2   4.20 s .
Thus:        vxf  3.10 m s .

P42 (7e-P44)
y  3.00t 3 : At t  2.00 s , y  3.00  2.00  24.0 m and: vy  dy  9.00t 2  36.0 m s  .
3

dt
If the helicopter releases a small mailbag at this time, the mailbag starts its free fall with velocity 36 m/s upward.
yb  ybi  vi t  gt 2  24.0  36.0t   9.80 t 2 . Setting yb  0 : 0  24.0  36.0t  4.90t 2 .
The equation of motion of the mailbag is:                        1                     1
2                     2
Solving for t, (only positive values of t count), t  7.96 s .

P43 (7e-P39)
(a) y f  yi  vi t  1 at 2 : 4.00   1.50 vi   4.90 1.502 and vi  10.0 m s upward .
2

(b) v f  vi  at  10.0   9.801.50  4.68 m s                v f  4.68 m s downward

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 3 posted: 10/4/2012 language: Latin pages: 3