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# Instantaneous Velocity and Acceleration for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

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)=2t3 – 10t2 + 5. Find (a) the position, (b) instantaneous velocity, (c) acceleration, and (d) speed of the particle at t =1.

Solution:

1. s (1) = 2(1)3 – 10(1)2 + 5 = –3
2. v(t) = s'(t) = 6t2 – 20t
3. v(1) = 6(1)2 – 20(1) = – 14

4. a(t) = v'(t)=12t – 20
5. a(1) =12(1) – 20 = – 8

6. Speed =|v(t)|=|v(1)|=14.

### Example 2

The velocity function of a moving particle is v(t) = – 4t2 + 16t – 64 for 0 ≤ t ≤ 7.

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

v(t)= – 4t2 +16t – 64
a(t) = v'(t) = t2 – 8t +16

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

1. The minimum acceleration occurs at t =4 and a(4)=0.
2. 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.

1. When is the acceleration 0?
2. When is the particle moving to the right?
3. When is the speed the greatest?

Solution:

1. 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.
2. For 2 < t < 4, v(t) > 0. Thus the particle is moving to the right during 2 < t < 4.
3. Speed = |v(t)| at t =1, v(t) = – 4.
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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