A body is moving along a straight line with a velocity which varies according to the equation v = 9 t2 + 2t, where v is in feet per second and t is in seconds. Find the expression for the distance as a function of time. A. 3t3 + 2t2 + C B. 3t3 + t2 + C C. 20t + C D. 18t + C

V (t) = 9t² + 2t

To find a function for distance as a function of time:

Using Calculus:

Take the integral of V (t)

This means find the anti-derivative (if v (t) is the derivative of d (t), what must d (t) be?)

d (t) = (1/3) * 9 * t ²⁺¹ + (1/2) * 2 * t¹⁺¹

d (t) = 3t³ + t² + C

Answer is B) 3t³ + t² + C.

Why did we add C at the end? The velocity function doesn't tell us anything about our initial distance, so we add C to account for wherever we started at t=0. This is also known as our Constant of Integration