Motion Mixed Review – Combined – Answer Key
Mrs. Travers PSII
Motion Test - Travers
Concept Review Questions (Hewitt) – ANS KEY
Distance and Displacement
1.
What is common about these two terms? What is the difference? Using a
sketch/drawing explain the difference between the two.
Ans. Both terms describe change in motion. Both are measured in length units.
Distance is a scalar. Displacement is a vector. Distance depends on the path.
Displacement does not (just measures start to finish).
Speed
2.
Velocity
3.
4.
Distinguish average speed from instantaneous speed.
Ans. ave. speed = total distance/total time; instantaneous speed is at one
point (like speedometer in car)
Distinguish between speed and velocity?
Ans. Speed is a scalar quantity; velocity is a vector quantity
If a car moves with a constant speed, can you say that it also moves with a
constant velocity? Give an example to support your answer.
Ans. No. Speed may be same, but if direction changes, velocity changes
(example - turning or circular motion)
Acceleration
5.
When you are most aware of motion in a moving vehicle – when it is moving
steadily in a straight line or when it is accelerating? If a car could move with
absolutely constant velocity (no bumps at all), would you be aware of motion?
Ans. When it is accelerating; No unless you had a different frame of
reference.
6.
Explain the motion of something when both the velocity and acceleration have the
same sign.
Ans. When velocity and acceleration have the same sign, the acceleration
adds to the velocity. So the object gets faster and faster in the direction of the
velocity (+ right, - left).
7.
Explain the motion of something when the velocity and acceleration have
different signs.
Ans. When velocity and acceleration have different signs, the acceleration
takes away from the velocity. So the object gets slower and slower in the initial
direction of the velocity until it becomes zero and then acceleration starts adding to
the velocity in the opposite direction (object turns around).
8.
What is the relationship between displacement and time for an object moving at
a constant acceleration?
Who is generally credited for determining
this relationship and what apparatus did he use to observe this relationship?
The displacement of an object undergoing constant acceleration is proportional to the
time of travel squared (d = 1/2at2). Although this relationship was derived
mathematically before Galileo, Galileo was the first to OBSERVE this relationship using
inclined planes where he could slow the motion down to observe it.
Free Fall
9.
What is meant by a freely falling object?
Ans. An object moving under the sole influence of the force of gravity (in
other words, gravity is the only force on it so it accelerates at 10 m/s/s.
There is no air resistance which is what we typically observe in our every
day lives.
10. If I toss a ball in the air and it comes back down to me, it appears to be
constantly moving. Is the velocity ever zero? What is its acceleration? Is it
constant?
Ans. Yes, when objects change direction, their velocities must go through
zero. So first the ball is going in the positive direction, and then in the negative
direction. It is constantly accelerating at a rate of 10 m/s/s in the negative
(downward direction). It is in free fall the whole time (we assume – no air resistance).
Problems.
1. 13.3 km/min
2. 4 km
3. 600 sec
4. distance: 26 blocks; displacement: 0
5. I: constant velocity (no a); II: constant positive acceleration, speeding up;
III: constant velocity (no a): IV: constant negative acceleration
(deceleration), slowing down.
Time
(sec)
0-2
2-4
4-6
6-9
Area
under
graph
(m)
12
18
24
18
Total
position
(m)
12
30
54
72
6. 3.84 x 105 km
7.
Make a table.
t (hours)
4
0.5
3
1.5
3
v (mph)
70.
0
65
0
80
12 hours
d (miles)
280
0
195
0
240
715 miles
Ave. Velocity = total distance/total time = 715 mi/12 hours = 60. Mph
8.
A) vf = 34.1 m/s
B) ∆x = 58.1 m
C) v ave = 17.0 m/s
9. 0.80 m/s2
10 a. Use vf2 = vi2 + 2aΔx; ∆x = 180. m
b. Use vf = vi + aΔt ; t = 10.21 se
11. ∆x = 45 m; t = 6 sec
12. t =0.70 sec; Hint: Calculate how long it would take the athlete to come down from
the peak height and double the time.