Question

The position of a particle moving along the x-axis is given by the function f(t)=4t3−39t2+108t . During which time interval(s) is the particle’s speed increasing?

Answer 1

To determine when the particle's speed is increasing, we need to find the intervals where the particle's velocity is positive and its acceleration is also positive.

First, let's find the particle's velocity function by taking the derivative of the position function:

f'(t) = 12t^2 - 78t + 108

Next, let's find the particle's acceleration function by taking the derivative of the velocity function:

f''(t) = 24t - 78

To find when the particle's speed is increasing, we need to determine when the velocity is positive and the acceleration is also positive.

Setting the velocity function greater than zero, we have:

12t^2 - 78t + 108 > 0

Factoring the quadratic, we get:

6(2t-9)(t-2) > 0

This quadratic is positive when either both factors are positive or both factors are negative.

Setting the factors greater than zero, we have:

2t - 9 > 0

t - 2 > 0

Simplifying, we have:

t > 4.5

t > 2

Since the second factor gives a wider range than the first, we can conclude that the velocity is positive when t > 4.5.

Next, setting the acceleration greater than zero, we have:

24t - 78 > 0

Simplifying, we have:

t > 3.25

Therefore, the particle's speed is increasing when t > 4.5 and t > 3.25.

Thus, the time interval when the particle's speed is increasing is t > 4.5.