HELP! Calculus question:?

Question

The height of a rocket at time t is 2x^3 + 7x^2 + 7x + 6 

Find the acceleration at time t = 2 
Answers: 
+36 
+37 
+38 
+39 

Answer 1

The rate of change in displacement is called velocity. It is denoted by v = ds/dt

The rate of change in velocity is called accelaration.It is denoted by a = dv/dt or (d²s)/(dt²).

In this case s = 2x³ + 7x² + 7x + 6

So the first derivative is

= 6x² + 14x + 7

v = 6x² + 14x + 7

And the second derivative is a = 12x + 14

when t = 2

a(t) = 12(2) + 14

= 24 + 14

= 38

Option 3 is coorect.