# Find the open t-intervals on which the particle is moving to the right. (Enter your answer using interval notation.)?

x(t) = t3 − 12t2 + 21t − 5, 0 ≤ t ≤ 10

(c) Find the velocity of the particle when the acceleration is 0.

(b) Find the open t-intervals on which the particle is moving to the right. (Enter your answer using interval notation.)

C) v(t) = 3(t − 1)(t − 5) , t=4 , v(t)=-9?

### 2 Answers

- electron1Lv 71 year ago
I will assume the unit of distance is meters.

x(t) = t^3 – 12 * t^2 + 21 * t – 5

The equation of velocity versus time is the first derivative of this equation.

v(t) = 3 * t^2 – 24 * t + 21

The initial velocity is 21 m/s. The equation of acceleration versus time is the first derivative of this equation.

a(t) = 6 * t – 24

6 * t – 24 = 0

t = 4 seconds.

Let’s use this time in the velocity versus time equation.

v = 3 * 2^2 – 12 * 2 + 21 = 9 m/s

Since the velocity is positive, I assume the particle is moving to the right. The next step is to determine the time when the particle’s velocity is 0 m/s.

3 * t^2 – 24 * t + 21 = 0

t = [24 ± √(576 – 4 * 3 * 21)] ÷ 6

t = [24 ± 18] ÷ 6

t1 = 7 seconds

t2 = 1 second

These are the two times when object’s velocity is 0 m/s. This is all that I know how to do.

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- Some BodyLv 71 year ago
x(t) = t³ − 12t² + 21t − 5

Velocity is the first derivative:

v(t) = 3t² − 24t + 21

Acceleration is the second derivative:

a(t) = 6t − 24

When acceleration is 0:

0 = 6t − 24

t = 4

Find v(4).

The particle is moving to the right when velocity is positive. First find where velocity is 0:

0 = 3t² − 24t + 21

0 = t² − 8t + 7

0 = (t − 1)(t − 7)

t = 1, 7

Evaluate v(t) on the intervals (0, 1), (1, 7), and (7, 10). Determine which ones are where v(t) is positive.

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