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# Vertical Motion and Horizontal Motion 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

### Vertical Motion

Example

From a 400-foot tower, a bowling ball is dropped. The position function of the bowling ball s (t)= – 16t2 +400, t ≥ 0 is in seconds. Find:

1. the instantaneous velocity of the ball at t =2 s
2. the average velocity for the first 3 s
3. when the ball will hit the ground

Solution:

1. v(t) = s '(t)= – 32t
2.  v(2)=32(2)= – 64 ft/s

3. Average velocity = = – 48 ft/s.
4. When the ball hits the ground, s (t) = 0.
5.  Thus, set s (t) = 0 – 16t2 + 400 = 0; 16t2 =400; t = ± 5.

Since t ≥ 0, t = 5. The ball hits the ground at t =5 s.

### Horizontal Motion

Example

The position function of a particle moving in a straight line is s (t) = t3 – 6t2 + 9t – 1, t ≥ 0. Describe the motion of the particle.

Step 1:  Find v(t) and a(t). v(t) = 3t2 – 12t + 9.

a(t) = 6t – 12

Step 2:  Set v(t) and a(t) = 0.

Set v(t) = 0 3t2 – 12t + 9 = 0 3(t2 – 4t + 3) = 0
3(t – 1)(t – 3) = 0 or t = 1 or t =3.
Set a(t) = 0 6t – 12 = 0 6(t – 2) = 0 or t = 2.

Step 3:  Determine the directions of motion. See Figure 9.3–3.

Step 4:  Determine acceleration. See Figure 9.3–4.

Step 5:  Draw the motion of the particle. See Figure 9.3–5.

s (0) = – 1, s (1) = 3, s (2) = 1 and s (3) = – 1

At t = 0, the particle is at –1 and moving to the right. It slows down and stops at t = 1 and at t = 3. It reverses direction (moving to the left) and speeds up until it reaches 1 at t = 2. It continues moving left but slows down and stops at –1 at t = 3. Then it reverses direction (moving to the right) again and speeds up indefinitely. (Note: "Speeding up" is defined as when |v(t)| increases and "slowing down" is defined as when |v(t)| decreases.)

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

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