How much distance?

Question

A particle starting at rest at t=0 moves along a line so that its acceleration at time t is 12t ft/sec^2. How much distance does the particle cover during the first 3 sec? 

 

Answer: 54 ft

 

Please show all your work.

Answer 1

Let s=distance.  The acceleration is

d²s/dt²=12t given, from which we get the velocity

ds/dt = 6t²+v₀, where v₀ is the initial velocity, given as 0.  The body starts at rest.  From which we get the position

s(t)=2t³+s₀, where s₀ is the initial position, given as 0

s(t)=2t³, which at t=3 equals 54