Particle Motion

Question

A particle moves along the x-axis with the velocity given by v(t) = 6t^2 + 2t for time t ≥ 0. If the particle is at position x = 5 at time t = 0 , what is the position of the particle at time t = 2 ?

Answer 1

v(t) = 6t² +2t

v(0) = 5

We need to first find the general form of the antiderivative of v(t)

s(t) = 6(1/3)t³ + 2(1/2)t² + C

s(t) = 2t³ + t² + C

Now we find the particular solution for s(0) = 5 by solving for C

5 = 2(0)³ + (0)² + C

5 = C

So our particular solution is

s(t) = 2t³ + t² + 5

Then we find s(2)

s(2) = 2(2)³ + 2² + 5

s(2) = 16 + 4 + 5 = 25