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 velocity,
Δd = Δt(v1 + v2) / 2
The position (displacement from the starting point) of the skier is determined at 1-s intervals and tabulated below. This was determined from the area under the curve, or from the above equation (which is the same thing). The corresponding position-time graph is shown to the right.
The position-time graph for constant
acceleration is curved (parabolic).
(See below for answers)
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 the following graph.
5. Sketch v-t and a-t graphs that correspond to the following 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 the graph below
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. Using the following 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
d. The acceleration of the object at
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.
For solutions to all the problems on this page click here.