One-Dimensional
Motion
Three easy steps to solving one-dimensional motion
problems by using kinematics equations:
1. Memorize the three must-be-memorized
kinematics equations for a constant
acceleration motion:
d:
displacement, 
:
initial velocity, 
:
final velocity, a: acceleration, and t: time
2. Construct a table as shown below and
fill in three variables’ data provided by the question itself.
|
|
|
d (m) |
a
(m/sec2) |
t (sec) |
|
|
|
|
|
|
Constant
acceleration motion problems always have 5 variables, which are: displacement (d), initial velocity (vi), final velocity (vf), acceleration (a), and time (t). The general rule is that once you know the values of three
variables, you should be able to get the rest of them by using the kinematics
equations.
3. Solve this problem by choosing one or
two of the kinematic equations according to the data available.
Examples:
1. If
a treadmill starts at a velocity of 2.5 meters per second and has a velocity of
1 meter per second after 2 minutes, what is the average acceleration of the
treadmill?
Answer:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
2.5 |
1 |
|
|
2
× 60 |
1 = 2.5 + a × 120
a = 
= −
0.0125 m/s2
2. With
an average acceleration of —1.0 m/s2,
how long will it take a cyclist to bring a bicycle with an initial velocity of
12.0 m/s to a complete stop?
Answer:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
12 |
0 |
|
—1.0 |
|
= 12 —1.0 × t
t
= 
= 12 s
3. A
helicopter is moving upward at a constant velocity of 20 m/s. A package is
released from the rising aircraft at a height of 225 m. How long does it take
for the package to reach the ground?
Answer:
It is a free fall so the package’s
acceleration is —9.8 m/s2 and its displacement is —225 m, negative
values because of going downwards.
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
20 |
|
—225 |
—9.8 |
|
—225 = 20 t — 
(9.8) t2
4.9 t2 —20t —225 = 0
t = 9.1 s
4. A
helicopter lifts off from the deck of an aircraft carrier and accelerates
upwards at a rate of 5 m/s2. The deck is 30 m above the surface of
the water. At 10 seconds into the flight, a coffee mug falls out of the door of
the helicopter. How long does it take for it to hit the water?
Answer: The falling of the mug is a free fall, so its acceleration is —9.8 m/s2.
We
need to find out what is the velocity and the position when the coffee mug
falls.
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
0(lift
off) |
|
|
5 |
10 |
vf = 0 + 5× 10 = 50 m/s (coffee mug’s velocity when
it falls.)
=
0 × 10 + × 5× 102 = 250 m
Coffee mug’s position is 30 + 250 = 280 m above the sea.
Now, for the coffee mug:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
50 |
|
—280 |
—9.8 |
|
=
50t — 
(9.8) t2
4.9t2 —50t —280 = 0
t = 14 s
It
needs 14 seconds for the coffee mug to reach the water.
5. A
ball rolls down an incline. It is released from rest and rolls 2 m in the first
second. How far does it roll in the third second?
Answer: This question asks that in the 3rd second, how far the ball
rolls. Therefore, we need to know the velocity of the ball at the beginning of
3rd second and its acceleration from the clues given above.
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
0 |
|
2 |
|
1 |
a = 4 m/s2
(after 2 seconds)
Now for the 3rd second:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
8 |
|
|
4 |
1 |
d = 8 × 1 + 4 × (1)2 = 10 m
In the 3rd second, it rolls 10 m.
Solve One-Dimensional Motion by Using
Graphs
If
you are able to graph position, velocity, or acceleration vs. time and apply
their relationships, you can solve all the SAT II problems of constant
acceleration motion.
Examples:
1. What
does the slope of a position versus time graph represent? What does the slope
of a velocity versus time graph represent?
Answer:
The slope of a position versus time graph represents object’s
velocity. The slope of a velocity versus time graph is the object’s
acceleration.
2. What
does the area between the curve and the time-axis represent on acceleration
versus time graph? What does the area between the curve and the x-axis
represent on a velocity versus time graph?
Answer:
On
the acceleration versus time graph, the area between the curve and the time-axis
is the changed velocity over the time periods. On a velocity versus time graph,
the area between the curve and the x-axis is the displacement over the time
periods.
3. A
late passenger, sprinting at 3 m/s, is 18 m away from the rear end of a train when
it starts out of the station with an acceleration of 0.25 m/s2.
a.
When will the person be next to the rear
of the train?
Answer:
When the person next to the train, the
person’s displacement is 18 m greater than that of the train.
Let t be the time needed to catch the train:
3t (person’s displacement) — 18 = (0.25) × t2 (train’s displacement)
t2 —24t + 144 = 0
t = 12 sec
b.
Solve this question by using a graph.
Answer:
So the area under the passenger is 3 × t
and the area under the train is
× t × 0.25t.
3t — 18 = 
t2
t = 12 sec
c.
What is the greatest acceleration the
train can have and still have the person reach the rear?
Answer: Let
the train’s acceleration be a, then the slope of the train from the graph above
is a.
The person’s total displacement minus 18 is less than train’s
displacement.
3t — 18 < × t × at
at2 —6t +36 > 0
(—6)2 — 4 × a × 36 < 0
If the train acceleration is greater than 1 m/s2,
this late passenger won’t be able to catch the train.
Algebra
review note: If 
is
always true, then both of b2 — 4ac < 0 and a > 0 must also be
true.
Free Fall
According
to Galileo’s gravity experiments, all objects that fall near the surface of
earth will have the same acceleration of 9.8 m/sec2. Since the free
fall acceleration is constant, we can apply kinematics equations, and the three
easy steps as discussed above, to solve all the free fall problems by setting its
acceleration equal to —9.8 m/sec2. A negative number indicates that
gravity is pointing downwards.
If
an object is thrown upward, the acceleration of the object is —9.8 m/s2,
constant at all times. The velocity of the object is 0 m/s when the object reaches its maximum height. The time upward is the same as downward, when the object is going
back to the original height (or position).
Examples:
1. A
rock climber, together with her first-aid kit, is descending a vertical rope at
a steady rate of 2.0 m/s. The strap holding the first-aid kit to her pack
breaks and the kit is “released”. After 2.5 seconds, the kit hits the ground.
How high above the ground is the first-aid kit when the strap breaks?
Answer:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
—2 |
|
|
—9.8 |
2.5 |
at2
d
= (—2) × 2.5 —0.5 × 9.8 × 2.52 = —35.6 m
2. A
ball is thrown upwards at a velocity of 20.0 m/s. How long will it take for the
ball to reach a height of 20.0 meters above the ground?
Answer:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
20 |
|
20 |
—9.8 |
|
at2
20 = 10t — (9.8)t2
4.9t2 —20t + 20 = 0
t = 2 sec
3. A
helicopter is traveling straight upwards at 6 m/s. When the helicopter is 100
meters above the ground, the pilot mistakenly lets go of his coffee mug and it
falls out of the window. How long does it take for the mug to hit the ground?
Answer:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
6 |
|
—100 |
—9.8 |
|
at2
4. The
acceleration of gravity on the moon is 1.6 m/s2 (downward). If you
can throw a ball upwards to a height of 20 meters on Earth, how high could you
throw it on the moon?
Answer:
On Earth:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
|
0 |
20 |
—9.8 |
|
vi
= 19.8 m/s
On Moon:
|
vi
(m/sec) |
vf
(m/sec) |
d
(m) |
a
(m/sec2) |
t
(sec) |
|
19.8 |
0 |
|
—1.6 |
|
5.
A person, staring through a 1.0-meter tall
window, sees a penny fall passed the window, and which takes 0.10 seconds to
pass. The person knows that the bottom of the window is 15 meters above the
ground. How tall is the building if the penny is dropped from the roof?
Answer: In order to find the distance from the
roof to the top of the window, we need to know the velocity of the penny at the
top of the window.
From the roof to the top of the window:
(
9.5)2
= 2(9.8) × d
d = 4.62 m
The height of the building equals: 4.62 + 1.0 + 15 = 20.61 m