3.4 POSITION, VELOCITY, AND ACCELERATION
Position: starting function usually denoted as x(t) or s(t)
Velocity:the rate of change for one’s position
- Average Velocity:
-This is the same as the average rate of change except for the change in notation. The change in notation is simply caused by relating the problem as a function of time.
- Instantaneous Velocity:
-Recognize that this is the derivative formula. Therefore the first derivative of a position function represents the velocity of the body in motion.
-Velocity represents not only the rate of change but the direction a particle is traveling.
a) (+) velocity = motion to the right or upward
b) (-) velocity = motion to the left or downward
Example 1
A dynamite blast blows a heavy rock upward with a launch velocity of 160 ft/sec. It reaches a maximum height of 160t – 16t2 ft after t seconds.
a) How high does the rock go?
b) How fast is the rock traveling when it is 256 ft above the ground?
c) Find the average velocity of the rock over the time interval 2 ≤ t ≤ 6
Speed: the absolute value of velocity – it does not take into account direction like the velocity does
Acceleration: the rate at which the velocity is changing. Since the derivative represents the rate of change and the acceleration is the rate of change of the velocity, then a(t) =
Therefore:
- (+) acceleration implies the velocity is increasing because there is a positive rate
of change
- (-) acceleration implies the velocity is decreasing because there is a negative rate
of change
Example 2
What was the acceleration of the rock in the example above? What was its acceleration after 2 seconds? Why is the acceleration (-)?
OTHER RULES TO KNOW ABOUT VELOCITY AND ACCELERATION
1)When the velocity and the acceleration of a particle have the same signs, the particle’s speed is increasing
2)When the velocity and acceleration of a particle have the opposite signs, the particle’s speed is decreasing
3) When the velocity is zero and the acceleration is not zero, the particle is changing
directions
Examples
1. If the position of a particle as a function of time t is given by x(t) = t3 – 11t2 + 24t, find the velocity and acceleration of the particle at time t = 5.
2. If the position of a particle is given by x(t) = t3 – 12t2 + 36t + 18, where t>0, find the point at which the particle changes direction.
3. Given the position function s(t) = t3 – 3t2 + 2t,
a) Find the body’s velocity and acceleration at the beginning and the end of the interval.
b) Find the average velocity of the body for the given interval.
4. Given that the position of a particle is found by x(t) = t3 – 6t2 + 1, t>0, find the displacement of the particle from t = 2 to t = 5.
5. The table below shows the times and distance of a runner. Use the table to answer a) – c)
Meters / 0 / 405 / 837 / 1024 / 1578min / 0 / 2 / 4 / 6 / 8
a) What is the average speed of the runner over the first 4 minutes? Include units
b) What is the approximate speed of the particle at t = 4 minutes? Include units
c) During what interval of the race is the runner moving fastest.
The graph shown below represents the velocity of a particle moving in a straight line on the open interval 0 < t < 10. Use the graph to answer questions 6-10.
30 ( meters/sec)
20
10
t (seconds)
-10 1 2 3 4 5 6 7 8 9 10
-20
-30
6. When does the particle change directions? Explain
7. When is the particle moving to the left? Explain
8. When is the particle moving to the right? Explain
8. When does the acceleration of the particle equal zero?
9. When is the particle speeding up? Explain
10. When is the particle slowing down? Explain