1. Displacement, Velocity and Acceleration

2. A stone is dropped from a 100m tall building
A stone is dropped from a 100m tall building. We can model the distance the stone travels from the top of the building as a function of time, or:
Complete the following table for the given times and graph it on the following graph: t d 1 2 3 4
4.9
19.6
44.1
78.4

3. A stone is dropped from a 100m tall building
A stone is dropped from a 100m tall building. We can model the distance the stone travels from the top of the building as a function of time, or: t d 1 2 3 4
4.9
19.6
44.1
78.4
What is the average velocity between 1 second and 3 seconds?

4. Velocity is the rate of change of the displacement, or distance, with respect to time. Therefore:
In addition, the rate of change of the velocity with respect to time is the acceleration. Therefore:

5. In addition, if:
then:
then:
Likewise, if:

6. s v a Differentiate with respect to time
Displacement Velocity Acceleration
s v a
Integrate with respect to time

7. So: Lastly, if: and: then: and:
Use this formula if you are given an acceleration or velocity function which is given in terms of position rather than time.

8. The velocity function is given below
The velocity function is given below. Find the acceleration in terms of its position.
Answer:

9. The displacement s meters of a body from an origin O at time t seconds is given by the equation below. Find the velocity and acceleration when
Answer:
A body moves in a straight line. At time t seconds its acceleration is given by the equation below. When t = 0, the velocity v of the body is 2 m/s and its displacement s from the origin O is 1 meter. Find expression for v and s in terms of t.
Answer:

10. (a) The values of t when the body is at rest.
If and s = 4 when t = 3, find:
(a) The values of t when the body is at rest.
(b) The acceleration when t = 5
(c) The displacement when t = 1
Answers: (a) 1, (b) (c) 5 1/3 m
If a = 1 – t and when t = 2, v = 1 and s = 4 2/3 find expressions for v and s in terms of t.
Answer:

11. If and, when t = 0, v = 9 and s = 6 find the values of t when the body is at rest and the displacement of the body from O at these times.
Answer: Body at rest at t = 1, s = 10 m and t = 3, s = 6 m
A body starts at rest at an origin O and its acceleration at time t seconds later is given by for and for Find an expression for s, the displacement from O at time t for and hence find s when t = 8.
Answer:
when t = 8, s = 182 m

12. A car travels at 80 km/hr for four hours. How far did it travel?
Let’s look at the definite integral of the velocity function.

13. Therefore, the area under a velocity/time graph represents the total distance traveled. This can be found by integrating the velocity function.

14. An automobile accelerates from rest at miles per hour for 9 seconds.
(a) What is the velocity after 9 seconds?
(b) How far does it travel in those 9 seconds?
Answer: (a) 63 miles per hour
(b) miles