Homework 3 Solutions
Chapter 3
Review Questions
1.
Q5: 15km
ൊ
0.5hrs. = 30km/h
2.
Q6: (25km/h)(0.5hrs.) = 12.5km
3.
Q7: Velocity has direction, speed only has magnitude
4.
Q8: Yes, but the opposite is NOT true
5.
Q9: Constant speed, but not constant velocity. The direction is changing and therefore
centripetal acceleration is present.
6.
Q11: Poorly written question (assume constant acceleration). (100km/h
‐
0km/h)
ൊ
(1h
ൊ
360) = 36000km/h
2
or (0.0278km/s
‐
0km/s)
ൊ 10s ൌ 0.00278km/s
7.
Q12: ሺ100km/h – 100km/hሻ ൊ
(1h
ൊ
360) = 0km/h
2
8.
Q13: Accelerating, no
9.
Q14: Linear motion
10.
Q15: The change in speed per given account of time was the same, therefore the
acceleration due to gravity is constant
11.
Q16: It is equal to the acceleration times time (v = at).
12.
Q17: Velocity was greater with a steeper incline. Therefore, the acceleration due to gravity
acts in the vertical direction.
13.
Q19: ~10.0 m/s
14.
Q20: (10.0m/s
2
)(5s) = 50m/s, (10.0m/s
2
)(6s) = 60m/s
15.
Q22: 10m/s
16.
Q24: d =
½
gt
2
= (0.5)(10.0m/s
2
)(1.0s
2
) = 5.0m, d =
½
gt
2
= (0.5)(10.0m/s
2
)(16.0s
2
) = 80m
Plug and Chug
17.
Q2: S
ave.
= 4m
ൊ
2s = 2 m/s
18.
Q4: S
ave.
= 24m
ൊ
0.5s = 48 m/s
19.
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 Summer '11
 Ramos
 Work, Christian cross, cross sectional area, Non‐zero initial velocity, acceleration times time

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