Question

[Calculus] analyze speed from position. Show steps, please!

Answer 1

Answer:

B

Step-by-step explanation:

We are given that the position of a particle is modeled by the function:

$s(t)=2t^3-24t^2+90t+7$

And we want to find the times for which the speed of our particle is increasing.

In other words, we want to find the times for which our acceleration is positive (a(t)>0).

So first, we will find our acceleration function. We can differentiate twice.

By taking the derivative of both sides with respect to t, we acquire:

$\displaystyle s^\prime(t)=v(t)=\frac{d}{dt}\big[2t^3-24t^2+90t+7\big]$

Differentiate:

$v(t)=6t^2-48t+90$

This is our velocity function. We can differentiate once more to acquire the acceleration function. Therefore:

$\displaystyle v^\prime(t)=a(t)=\frac{d}{dt}\big[6t^2-48t+90\big]$

Differentiate:

$a(t)=12t-48$

If our speed is increasing, our acceleration must be positive. So:

$a(t)>0$

By substitution:

$12t-48>0$

Now, we can solve for t:

$12t>48\Rightarrow t>4$

Therefore, the only interval for which the speed of the particle is increasing (i.e. the acceleration is positive) is for all times t>4.

So, our answer is B.