Practice problems for these concepts can be found at: More Applications of Derivatives Practice Problems for AP Calculus

Position Function: *s (t)*

Instantaneous Velocity: *v(t)* =*s'(t)* =

If particle is moving to the right →, then *v(t)* > 0.

If particle is moving to the left ←, then *v(t)* < 0.

Acceleration: *a(t)* = *v'(t)* = or *a(t)* = *s''(t)* =

Instantaneous speed: |*v(t)*|

### Example 1

The position function of a particle moving on a straight line is *s (t)*=2*t*^{3} – 10*t*^{2} + 5. Find (a) the position, (b) instantaneous velocity, (c) acceleration, and (d) speed of the particle at *t* =1.

Solution:

*s*(1) = 2(1)^{3}– 10(1)^{2}+ 5 = –3*v(t)*=*s'(t)*= 6*t*^{2}– 20*t**a(t)*=*v'(t)*=12*t*– 20- Speed =|
*v(t)*|=|*v*(1)|=14.

*v*(1) = 6(1)^{2} – 20(1) = – 14

*a*(1) =12(1) – 20 = – 8

### Example 2

The velocity function of a moving particle is *v(t)* = – 4*t*^{2} + 16*t* – 64 for 0 ≤ *t* ≤ 7.

What is the minimum and maximum acceleration of the particle on 0 ≤ t ≤ 7?

*v(t)*= – 4

*t*

^{2}+16

*t*– 64

*a(t)*=

*v'(t)*=

*t*

^{2}– 8

*t*+16

See Figure 9.3-1. The graph of *a(t)* indicates that:

- The minimum acceleration occurs at
*t*=4 and*a*(4)=0. - The maximum acceleration occurs at
*t*=0 and*a*(0)=16.

### Example 3

The graph of the velocity function is shown in Figure 9.3-2.

- When is the acceleration 0?
- When is the particle moving to the right?
- When is the speed the greatest?

Solution:

*a(t)*=*v'(t)*and*v'(t)*is the slope of tangent to the graph of*v*. At*t*=1 and*t*=3, the slope of the tangent is 0.- For 2 <
*t*< 4,*v(t)*> 0. Thus the particle is moving to the right during 2 <*t*< 4. - Speed = |
*v(t)*| at*t*=1,*v(t)*= – 4.

Thus, speed at *t* = 1 is |–4| = 4 which is the greatest speed for 0 ≤ *t* ≤ 4.

Practice problems for these concepts can be found at: More Applications of Derivatives Practice Problems for AP Calculus

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