calc question is below~~!!!! help me please?

Question

the acceleration of a particle moving along the x-axis at time t is given by a(t)=4t-12. If the velocity is 10 when t=0and the position is 4 when t=0, then the particle is changing direction at???

the choices are:
a) t=1
b) t=3
c) t=5
d) t=1 and t=5
e) t=1 and t=3 and t=5

Answer 1

Well we know the formula for x''(t) = 4t - 12
All we need to do is find the antiderivatives and use the two data points to fill in the unkowns.

if x''(t) = 4t - 12, then
x'(t) = K + 4t^2/2 - 12t = K+2t^2 - 12t
Plugging in the value of 10 when t is zero, we have
x'(0) = 10 = K
so x'(t) = 10 + 2t^2 - 12t

then x(t) = L + 10t + 2t^3/3 - 12t^2/2 = L + 10t - 6t^2 + 2/3 t^3
We are given x(0) = 4 so L = 4
But I'm not sure the last step is needed, since we only want to know when the particle is changing direction, which is when x'(t) = 0.
So find the roots of 2t^2 - 12t + 10, or t^2 - 6t + 5 = (t-5)(t-1)
or t=5 and t = 1