Answer to Question #122958 in Calculus for Gagandeep Kaur

Question

a) The equation of motion of a particle is ,where s=t^3-3t is in metres and t is in seconds. Find i) the velocity and acceleration as functions of t, ii) the acceleration after 2 seconds, iii) the acceleration, when the velocity is 0.

Answer 1

 $S=t^3-3t ~~~~~ \text{ (Given)}\\[1 em] \text{ Velocity is rate of change of position with respect to time.}\\[1 em] v(t)=\frac{ds}{dt}=3t^2-3\\[1 em] \text{ Acceleration is the rate of change of velocity respect to time.}\\[1 em] a(t)=\frac{dv}{dt}=6t\\[1 em] \text{ The acceleration after 2 seconds}=a(2)=6(2)=12 ~m /s^2 \\[1 em] \text{ when the velocity = 0 }\\[1 em] \therefore v(t)=3t^2-3=0 \\[1 em] \therefore 3t^2=3 \implies t^2=1\therefore t=1 ,t=-1 \\[1 em] \text{ We reject the negative answer because time cannot be negative}.\\[1 em] \text{ Now substitute in the acceleration at } t=1 \\[1 em] a(t)=6t\implies\Rightarrow a(1)=6(1)=6~ m /s^2$