Abody 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