At t=0 , a particle starts at the origin with a velocity of 6 feet per second and moves along the x-axis in such a way that at time t its acceleration is 12t^2 feet per second per second. Through how many feet does the particle move during the first 2 seconds?

Answer 1

a = 12t^2
velocity = 4t^3 + c
when t = 0, v = 6
6 = 4(0) + c
c = 6
v = 4t^3 + 6
distance = t^4 + 6t + k
when t = 0, distance = 0 ---> k = 0

distance = t^4 + 6t
so when t = 2
distance = 2^4 + 6(2) = 28 feet

Answer 2

Thanks, Reiny!