Physics help from experts.
This page introduces simple velocity and acceleration and the use of motion graphs, followed by practice problems.
Suppose a student is strolling down the street in a straight line at a steady speed. The student takes one pace every second and covers 0.80 m with each pace.
After 1 s the student has traveled 0.8 m.
After 2 s the student has traveled another .8m for a total of 1.6 m.
. . . and so on.
The data points form a straight line on a position-time graph.
Using any two points on the line results in a slope of 0.80 m/s.
The position-time graph for constant velocity is a straight line
The slope of a position-time graph is the velocity of the object.
Expressed as an equation, v = Δd/Δt .
Since at any time the velocity is + 0.8 m/s, it is easy to construct a velocity-time graph of the motion.
The velocity-time graph for constant velocity is a horizontal line.
The area of a rectangle is given by A = bh where b is the base and h is the height.
If you find the area under the curve of this particular graph for any 1 second interval you will find it to be
A = bh = (1 s)(0.80 m/s) = 0.80 m
The area under the curve for any 2 s interval is
A = bh = (2 s)((0.80 m/s) = 1.6 m
The area under a velocity-time graph gives the displacement during that interval, or,
Δd = vΔt
The student realizes she dropped something. She takes 3 paces backward, covering 1.0 m with each pace, and taking 1 pace each second.
The velocity during this interval is (-2 m)/(2 s) = -1 m/s.
On a position-time graph, a negative slope corresponds to a negative velocity. If forward is positive, then backward is negative.
The corresponding velocity-time graph shows a velocity of 0.80 m/s for the first 5 s, and -1.0 m/s for the last 3 s.
To use the velocity-time graph to find the displacement for the 8 s trip, add the areas under the curve:
(5 s)(0.80 m/s) + (3 s)(-1.0 m/s) = 4.0 m - 3.0 m = 1.0 m
This corresponds to the final position in the position-time graph, as it should.
Suppose a skier, starting from rest, goes down a hill in a straight run, picking up speed steadily.
At 1 s the skier's speed is .50 m/s, at 2s 1.0 m/s, at 3 0 s 1.5 m/s and so on for 5 s.
The velocity-time graph for constant acceleration is a straight line.
Using any two points on the line gives a slope of 0.50 m/s/s. This is the rate of change of velocity, or acceleration. In this case the skier gains 0.50 m/s each second.
the slope of a velocity-time graph is acceleration. Or,
a = Δv / Δt .
As before, the area under the velocity-time graph gives the displacement of the object for the specified time interval.
The area of a trapezoid is given by
A = b(h1 + h2) / 2 where b is base and h is height.
In this example the displacement of the skier from 2 s to 4 s is
(2 s)(1 m/s + 2 m/s ) / 2 = 3.0 m
Note that the heights are velocities, and the base is time interval.
In general, for constant acceleration,
Δd = Δt(v1 + v2) / 2
The position (displacement from the starting point) of the skier is determined at 1-s intervals and tabulated. This can be determined from the area under the velocity-time graph, or from the above equation (which is the same thing). The corresponding position-time graph is shown.
The position-time graph for constant acceleration is curved (parabolic).
|t (s)||d (m)|
1. Sketch corresponding d-t and v-t graphs for the motion of a car
a. at rest
b. moving forward with constant velocity
c. moving backward with constant velocity.
2. Sketch corresponding d-t and v-t graphs for the motion of a car whose acceleration is negative.
3. Sketch d-t, v-t, and a-t graphs for a ball thrown straight into the air, until it returns to the point of release.
4. Sketch v-t and a-t graphs that correspond to this graph.
5. Sketch v-t and a-t graphs that correspond to this graph.
6. Sketch d-t, v-t, and a-t graphs that depict the following motion. A ball starting from rest rolls down a ramp. At the bottom of the ramp it hits a spring which compresses, then sends the ball back up the ramp to the same point from which it was released.
7. Using this graph:
a. find the average velocity of the object for the interval from 4.0 s to 8.0 s.
b. find the instantaneous velocity at 8.0 s.
8. Using this graph, find
a. The displacement of the object from 2.0 s to 4.0 s.
b. The displacement of the object from 8.0 s to 10.0 s.
c. The acceleration of the object at 3.0 s.
d. The acceleration of the object at 10.0 s
a. Sketch the velocity time graph for a camel slowing down at a uniform rate moving in the negative direction.
b. Sketch a velocity time graph for an iguana slowing down at a uniform rate moving in the positive direction.