Answer:
A particle will be rest firstly at t = 3 sec and again t = 4 sec.
Explanation:
Given that,
The equation of position,
\(x(t)=2t^3-21t^2+72t-53\)
We need to calculate the velocity
On differentiating equation of position
\(v=\dfrac{dx(t)}{dt}=6t^2-42t+72\)
A particle is at rest when the velocity equal to zero.
So, \(6t^2-42t+72=0\)
\(t^2-7t+12=0\)
After solution,
\(t = 3\ sec\) , \(t=4\ sec\)
Hence, A particle will be rest firstly at t = 3 sec and again t = 4 sec.
