Question

a particle moves along the x-axis so that at time t >(or equal to) 0 its position is given by x(t)=2t^3 - 21t^2 + 72t-53. At what time t is the particle at rest? ...?

Answer 1

Particle is at rest when Velocity = 0
V = X ' ( t ) = 6t^2 - 42t + 72 = 0
V = t^2 - 7t + 12 = 0
Solve for t:t1 = 3t2 = 4Particle is at rest at t = 3 and t = 4.

Answer 2

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.