Answer:
C
Step-by-step explanation:
Remember that if s(t) is a position function then:
[tex]v(t)=s'(t)[/tex] is the velocity function and
[tex]a(t)=s''(t)[/tex] 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:
[tex]s(t)=t^2+4t+10[/tex]
So, let's take the first derivative first with respect to t:
[tex]\frac{d}{dt}[s(t)]=\frac{d}{dt}[t^2+4t+10][/tex]
Expand on the right:
[tex]s'(t)=\frac{d}{dt}[t^2]+\frac{d}{dt}[4t]+\frac{d}{dt}[10][/tex]
Use the power rule. Remember that the derivative of a constant is 0. So, our derivative is:
[tex]v(t)=s'(t)=2t+4[/tex]
This is also our velocity function.
To find acceleration, we want to second derivative. So, let's take the derivative of both sides again:
[tex]\frac{d}{dt}[s'(t)]=\frac{d}{dt}[2t+4][/tex]
Again, expand the right:
[tex]s''(t)=\frac{d}{dt}[2t]+\frac{d}{dt}[4][/tex]
Power rule. This yields:
[tex]a(t)=s''(t)=2[/tex]
So, our answer is C.
And we're done!
