Question

Solve the problem. If s is a distance given by s(t) = + + 40 + 10. find the acceleration, alt). Q ald) = 10 ald) = 2t + 4 O alt) = 2 O ald) = 2t​

Answer 1

Answer:

C

Step-by-step explanation:

Remember that if s(t) is a position function then:

$v(t)=s'(t)$ is the velocity function and

$a(t)=s''(t)$ is the acceleration function.

So, to find the acceleration, we need to solve for the second derivative of our original function. Our original function is:

$s(t)=t^2+4t+10$

So, let's take the first derivative first with respect to t:

$\frac{d}{dt}[s(t)]=\frac{d}{dt}[t^2+4t+10]$

Expand on the right:

$s'(t)=\frac{d}{dt}[t^2]+\frac{d}{dt}[4t]+\frac{d}{dt}[10]$

Use the power rule. Remember that the derivative of a constant is 0. So, our derivative is:

$v(t)=s'(t)=2t+4$

This is also our velocity function.

To find acceleration, we want to second derivative. So, let's take the derivative of both sides again:

$\frac{d}{dt}[s'(t)]=\frac{d}{dt}[2t+4]$

Again, expand the right:

$s''(t)=\frac{d}{dt}[2t]+\frac{d}{dt}[4]$

Power rule. This yields:

$a(t)=s''(t)=2$

So, our answer is C.

And we're done!