Expert: Richard J. Raridon - 11/12/2006

Question
Find the distance function s(t) if the acceleration function is given by the following formula, and if the initial velocity v0 and the initial distance s0 at time t=0 are known:
A) a(t)= kt where k is a constant
B) a(t)= square root of (2t + 1)
C) a(t)= sin (5t)
D) a(t)= sin (0.1t)/ cos^3 (0.1t)

I know that you have to find the second antiderivative of the acceleration function to find the distance funtion. i'm confused about the initial velocity v0 and why it is mentioned in this problem.

Answer
whenever you integrate without limits you have to add a constant and that's why you have an initial velocity.  
A) a(t) = kt, v(t) = (1/2)kt^2 +c,
s(t) = (1/6)kt^3 +ct +d where c and d are constants
C) v(t) = (-1/5)cos(5t) +c, s(t) = (-1/25)sin(5t) +ct +d