Question

A particle moves along the x-axis so that at time t >= 0 (t is greater than or equal to "t") its position is given by x(t)=-t^3+11t^2-24t. Determine the velocity of the particle at t=7.

Answer 1

$x(t) = -t^3 +11t^2 -24t \\ v(t) = xʼ(t) = -3t^2+22t -24$

At t = 7

$v(7) = -3 \times 7^2 + 22 \times 7 -24 \\ = -147 +154 -24 \\ = -17$

Answer: -17