- A particle moves according to the law of motion, $s= t^3 -12t^2 +36$, $t$ greater than or equal to $0$ . where ‘$t$’ is measure in seconds and $s$ is measured in metres.
a) Find the velocity of the partice after ‘$t$’ seconds.
b) What is the velocity after $4$ seconds?
c) What is the particle at rest?
d) When is the particle moving forward?
e) Find the total distance travelled by the particle during the first $8$ seconds.
Draw a diagram to illustrate the motion of the particle. where ‘$t$’ is measure in seconds and is measured in metres.
hint:
a. $v(t) = s'(t) = 3t^2 - 24t$
b. $v(4) = ....$
c. $v(t) = 0 \implies 3t^2 - 24t = 3t(t-8) = 0\implies t = ...$
d. The moving forward means $v \ge 0$. Thus solve: $3t(t-8) \ge 0 \implies t \ge 8$.
e. Total Distance $ = \displaystyle \int_{0}^8 (3t^2 - 24t) dt = ....$