Practice Problems

1.

inches
(1' = 12") / centimeters
(1" = 2.54 cm) / meters
(1 m = 100 cm)
5' x 12''/1'
= 60 " + 4" = 64" / 64" x 2.54 cm/1"
= 160 cm / 160 cm x 10-2 m/cm
= 1.6 m

2. a.

Mass / 48.307 g / 49.886 g / 50.911 g / 49.524 g
Mean / (48.307 + 49.886 + 50.911 + 49.525) = 49.657 g
4
Deviation / 1.350 g / 0.229 g / 1.254 g / 0.133 g
Average Deviation / (1.350 +0.229 + 1.254 + 0.133) = 0.742
4
% D / 100(0.742/49.657) = 1.49 %

b.

(100)(50.000 – 49.657)/50.000 = 0.686 %

c.

The measurements were more accurate than precise, because the % D was smaller than the % D.

3. a.

Fs (N)
30 /
20
10
0
0.1 / 0.2 / 0.3 / 0.4
x (m)

b.

slope = (32.0 – 0)/(0.40 – 0) = 80 N/m

c.

k = slope = 80 N/m

d.

area = ½bh = ½(0.40)(32.0) = 6.4 N•m

e.

W = area = 6.4 N•m

4.

x = rcosq = (110 km)cos25o = 100 km
y = rsinq = (110 km)sin25o = 46 km

5.

B / (15 mi + 25 mi)/1 hr = 40 mph

6.

A / (30 mi + 30 mi)/(1 hr + 0.6 hr) = 37.5 mph

7.

A / Velocity and displacement always go together, but acceleration is Dv, which could be from + to – v.

8.

D / you can have a constant velocity with displacement when acceleration is zero.

9.

C / Negative velocity is going backwards, but positive acceleration is opposite velocity \ slowing down.

10.

B / Displacement and velocity have the same sign, but gravity is always downward (–).

11.

D / Acceleration is constant and downward at all times

12.

B / Ball is momentarily stopped, but still's falling down.

13.

C / Same because both accelerate by gravity.

14.

C / Bill's ball is –v when it travels downward from h \ they both land with the same velocity.

15.

B / Ball goes from +v ® 0 ® -v (on the way down).

16.

C / The first ball is always faster \ it separates from the second.

17.

B / Same acceleration \ Dv is same, even though first is faster.

18. a.

7 m + 3 m = 10 m

b.

7 m + (-3 m) = 4 m

c.

vav = 4 m/10 s = 0.4 m/s

d.

speed = 10 m/10 s = 1 m/s

19. a.

d = x – xo = 30 m – 50 m = -20 m

b.

v = d/t = -20 m/2 s = -10 m/s

20. a.

d / vav / t
150 m / v / 17 s

b.

v = d/t = 150 m/17 s = 8.8 m/s

21.

d / vav / t
d / 5 m/s / 2 x 3600 s
d = vt = (5 m/s)(7200 s) = 36,000 m

22. a.

a = (vt – vo)/t = (15 m/s – 0 m/s)/5.0 s = 3 m/s2

b.

a = (vt – vo)/t = (5 m/s – 15 m/s)/5 s = -2 m/s2

23. a.

d / vo / vt / a / t
d / 15 m/s / 0 m/s / a / 3 s

b.

vt = vo + at
0 m/s = 15 m/s + a(3 s) \ a = - 5 m/s2


24.

d / vo / vt / a / t
d / 0 / 28 m/s / 2 m/s2 / t
vt2 = vo2 + 2ad
(28 m/s)2 = 0 + (2)(2 m/s2)d \ d = 196 m

25.

d / vo / vt / a / t
-100 m / 0 / vt / -10 m/s2 / t
d = vot + ½at2
100 m = 0 + ½(-10 m/s2)t2 \ t = 4.5 s

26.

d / vo / vt / a / t
d / 0 / 900 m/s / a / 180 s
d = ½(vo + vt)t = ½(0 m/s+ 900 m/s)(180 s) = 81,000 m

27.

d / vo / vt / a / t
d / 0 / vt / -10 m/s2 / 5 s
d = vot + ½at2 = 0 + ½(-10 m/s2)(5 s)2 = 125 m

28.

d / vo / vt / a / t
d / 10 m/s / 30 m/s / a / 10 s

a.

vt = vo + at
30 m/s = 10 m/s + a(10 s) \ a = 2 m/s2

b.

d = ½(vo + vt)t = ½(10 m/s+ 30 m/s)(10 s) = 200 m

29.

d / v / t
d / 30 m/s / 0.5 s

a.

d = vt = (30 m/s)(0.5 s) = 15 m
d / vo / vt / a / t
d / 30 m/s / 0 / -6.0 m/s2 / t

b.

vt2 = vo2 + 2ad
02 = (30 m/s)2 + (2)(-6.0 m/s2)d \ d = 75 m

c.

dtotal = d1 + d2 = 15 m + 75 m = 90 m

30. a.

vav = ½(vo + vt)
20 m/s = ½(0 + vt) \ vt = 40 m/s
d / vo / vt / a / t
d / 0 / 40 m/s / 2.5 m/s2 / t

b.

vt = vo + at
40 m/s = 0 + (2.5 m/s2)t \ t = 16 s

c.

d = vot + ½at2
d = 0 + ½(2.5 m/s2)(16 s)2 = 320 m

31. a.

on the way up / at the ball's highest / on the way down
-10 m/s2 / -10 m/s2 / -10 m/s2

b.

d / vo / vt / a / t
d / 20 m/s / 0 / -10 m/s2 / t

(1)

vt2 = vo2 + 2ad
0 = (20 m/s)2 + 2(-10 m/s2)d \ d = 20 m

(2)

vt = vo + at
0 = 20 m/s + (-10 m/s2)t \ t = 2 s

c.

How much time is the ball in the air? / 4 s
How fast is the ball when it returns? / -20 m/s

32.

A / position-time graph is curved.

33.

C / Their positions were equal at 0 and 7 seconds \ they have the same displacement.

34.

A / The slope of the line at 7 seconds is greater for car A.

35.

C / Velocity goes from 0 to –v.

36.

B / Velocity goes from +v ® 0 ® -v.

37.

v

0 t

38. a.

0 s to 8 s / 10 s to 18 s / 20 s to 24 s
v = d/t = 12 m/8 s
v = 1.5 m/s / v = d/t = 0 m/8 s
v = 0 m/s / v = d/t = -10 m/4 s
v = -2.5 m/s

b.

8 s to 10 s / 18 s to 20 s
a = Dv/t = -1.5 m/s/2 s
a = -0.75 m/s2 / a = Dv/t = -2.5 m/s/2 s
a = -1.25 m/s2

c.

v (m/s)

2
1
0
-1
-2
-3
4 / 8 / 12 / 16 / 20
t (s)

39. a.

0 s and 5-7 s

b.

Between 10 s and 11 s; this is when +Area = -Area

c.

0 to 2 s / 2 s to 3 s
a = Dv/t = (-1 m/s – 0)/2 s
a = -1 m/s2 / a = Dv/t = (0)/2 s
a = 0
3 s to 5 s / 5 s to 7 s
a = Dv/t = (0 – -2 m/s)/2 s
a = 1 m/s2 / a = Dv/t = (0)/2 s
a = 0
7 s to 11 s / 11 s to 12 s
a = Dv/t = (5 m/s – 0)/5 s
a = 1 m/s2 / a = Dv/t = (0)/1 s
a = 0

d.

a (m/s2)

1
0
-1
2 / 4 / 6 / 8 / 10
t (s)

40. a.

vt = vo + at
20 m/s = 0 + (2.5 m/s2)t \ t = 8 s

b.

d = vot + ½at2 = 0 + ½(2.5 m/s2)(8 s)2 = 80 m

c.

vt = vo + at
0 = (20 m/s) + (-4 m/s2)t \ t = 5 s

d.

d = ½(vo + vt)t = ½(20 m/s + 0)(5 s) = 50 m

e.

d = 200 m – 80 m – 50 m = 70 m
t = d/v = 70 m/20 m/s = 3.5 s

f.

ttotal = 0.3 s + 8 s + 3.5 s + 5 s = 16.8 s

41.

vt2 = vo2 + 2ad
(0)2 = (90 m/s)2 + 2a(605 m) \ a = -6.7 m/s2

42. a.

vav = ½(vo + vt)
40 m/s = ½(30 m/s + vt) \ vt = 50 m/s

b.

vt = vot + at
50 m/s = 30 m/s + (0.8 m/s2)t \ t = 25 s

c.

d = ½(vo + vt)t = ½(30 m/s + 50 m/s)(25 s) = 1000 m

43. a.

on the way up / at the highest point / on the way down
down / down / down

b.

vt2 = vo2 + 2ad
0 = (40 m/s)2 + 2(-10 m/s2)d \ d = 80 m

c.

vt = vo + at
0 = 40 m/s + (-10 m/s2)t \ t = 4 s

d.

2 x 4s = 8 s

e.

-40 m/s

44.

motion graphed / d vs. t / v vs. t / a vs. t
stationary / d / d / d
constant positive velocity / c / a / d
constant positive acceleration / b / c / a


45. a.

a (m/s2)

1
0
-1
2 / 4 / 6 / 8 / 10
t (s)

Practice Multiple Choice

1.

D / vt2 = vo2 + 2ad
vt2 = 0 + 2(5 m/s2)(250 m)\ vt = 50 m/s

2.

C / vt = vo + at
vt = (0.5 m/s) + (10 m/s2)(0.70 s) = 7.5 m/s

3.

A / vt = vo + at
11.2 m/s = 9.6 m/s + a(4 s) \ a = 0.4 m/s2

4.

C / vt2 = vo2 + 2ad
vt2 = 0 + 2(10 m/s2)(20 m) \ vt = 20 m/s

5.

D / d = ½(vo + vt)t
d = ½(0 + 30 m/s)(6 s) = 90 m

6.

A / Displacement is shortest distance between two points with changes direction displacement is less.

7.

D / d = vot + ½at2
d = 0 + ½(10 m/s2)(6 s)2 = 180 m

8.

A / d = vot + ½at2
40 m = 0 + ½(10 m/s2)t2 \ t = 2Ö2 = 2.8 s

9.

C / vt2 = vo2 + 2ad
0 = (30 m/s)2 + 2(-10 m/s2)d \ d = 45 m

10.

A / Acceleration is the change in velocity/time \ velocity increases by 9.8 m/s each second.

11.

D / At the highest point, it momentarily stops (v = 0), but acceleration due to gravity is constant (a < 0).

12.

C / d = vot + ½at2
3 m = 0 + ½a(1 s)2 \ a = 6 m/s2

13.

A / A position/time graph that is a straight diagonal line represents constant velocity \ the acceleration is zero.

14.

B / d = area of a triangle is ½ bh \
d = ½(t)(vt - vo) = ½(4 s)(20 m/s) = 40 m

15.

A / d = ½at2 \ curved line with increasing slope.

16.

D / Falling object has constant, negative acceleration.

17.

D / d = ½at2 \ curved upward, v = at \ diagonal line

18.

B / d = [(400 m)2 + (300 m)2]½ = 500 m
t = 60 s + 40 s = 100 s \ v = d/t = 500 m/100 s = 5 m/s


19.

C / When the area above and below the x-axis are equal (displacement = 0) is between 1.5 s and 2.0 s.

20.

A / Displacement equals area. Since the area under Car X is greater than under car Y, then car X is ahead.

21.

D / Displacement equals area. The area under the car Y equals the area under the car X at 40 s.

22.

C / d1 = vot + ½at2 ® 1 = 0 + ½a(1)2 \ a = 2 m/s2
d2 = vot + ½at2 = 0 + ½(2 m/s2)(2 s)2 = 4 m – 1 m = 3 m

23.

A / Object moving at constant velocity from 0 s to 1 s, slowing from +v to –v from 1 s to 2 s, then back.

24.

C / d = vot + ½at2 = (24 m/s)(11 s) + ½(-6 m/s)(10 s)2 = –36 m

25.

B / vav = d/t = 8 m/2 s = 4 m/s

26.

D / When accelerating from rest with constant acceleration, vt = 2vav \ vt = 2(4 m/s) = 8 m/s

Practice Free Response

1.

Cart is moving away from origin / 0-5 s, 9-11s
Cart is stationary / 0 s, 5 s, 11 s
Cart is moving toward the origin / 5-9 s, 11-12 s
Cart has returned to origin / 9 s
When farthest from the origin / 5 s
Positive acceleration / 0-2 s, 9-12 s
Zero acceleration / 2-3 s, 7-9 s
Negative acceleration / 3-7 s

2.

vt2 = vo2 + 2ad
(25 m/s)2 = 0 + 2a(180 m) \ a = 1.7 m/s2
vt2 = vo2 + 2ad
vt2 = (25 m/s)2 + 2(1.7 m/s2)(95 m) \ vt = 31 m/s

3. a.

d = vot + ½at2
d = 0 + ½(30 m/s2)(2 s)2 = 60 m

b.

vt = vo + at
vt = (30 m/s2)(2 s) = 60 m/s

c.

vt2 = vo2 + 2ad
0 = (60 m/s)2 + 2(-10 m/s2)d \ d = 180 m + 60 m = 240 m

d.

v = vo + at
0 = 60 m/s + (-10 m/s2)t \ t = 6 s + 2 s = 8 s

e.

d = vt
-240 m = (5 m/s)t \ t = 48 s


4. a.

Time, t (s) / 0.00 / 0.50 / 1.00 / 1.50 / 2.00
Position, x (m) / 0.00 / 0.22 / 0.70 / 1.35 / 2.10
Velocity, vav (m/s) / 0.44 / 0.96 / 1.30 / 1.50

b.

vt (m/s)

1.4
1.0
0.8
0.4
0
0.4 / 0.8 / 1.2 / 1.6
t (s)

c. (1)

slope = a = Dv/Dt
a = (1.60 – 0.60 m/s)/(1.80 – 0.40 s) = 0.71 m/s2

(2)

y-intercept = vo = 0.30 m/s

d. (1)

vt = vo + at
vt = 0.30 m/s + (0.71 m/s2)(2.00 s) = 1.7 m/s

(2)

vt2 = vo2 + 2ad
(1.7 m/s)2 = 0 + 2(0.71 m/s2)d \ d = 2.0 m

5. a.

vt2 = vo2 + 2ad = 0 + 2(-10 m/s2)(-15 m) = 17.3 m/s
vt2 = vo2 + 2ad \ 0 = (17.3 m/s)2 + 2a(1 m) \ a = -150 m/s2

b.

You would loosen the net so that the distance for deceleration is greater (v2 = vo2 + 2ad).