acceleration and distance

Question

A particle moving along the x-axis has acceleration given by . If the particle is initially at rest at the coordinate x = 3, what distance does it move in its first 4 seconds of motion?

Answer 1

a = dv/dt so:

dv = a dt

since a(t) = 2t then:

dv = 2t dt

∫dv = ∫2t dt

v(t) = t² + C

Since the particle is initially at rest, that means v(0) = 0 so C = 0

So v(t) = t²

But v = dx/dt so:

dx/dt = t²

dx = t² dt

∫dx = ∫t² dt

x(t) = t³/3 + C

We are told that at t = 0, x = 3 so t(0) = 3 therefore:

3 = 0³/3 + C or C = 3

So x(t) = t³/3 + 3

To find the distance at t = 4, plug it in:

x(4) = 4³/3 + 3 = 73/3 so the particle is at the x = 73/3 mark. BUT it started at x = 3 so its total distance traveled is 73/3 - 3 = 64/3 (units)