Question

The position of a particle moving along x axis given by x= -2t3+3t2+5. The acceleration of particle at the instant its velocity become zero is

Answer 1

thus the ans is -6 m/s^2

Answer 2

Answer:

The acceleration is 6 m/s² and -6 m/s²

Explanation:

Position of the particle is given by

$x=(-2t^3+3t^2+5)$ m

We know that rate of change of position (displacement) is velocity

Thus, the velocity is

$\boxed{v=\frac{dx}{dt}}$

$\implies v=\frac{d}{dt}(-2t^3+3t^2+5)$

$\implies v=-6t^2+6t$

If velocity is zero then

$v=0$

$\implies -6t^2+6t=0$

$\implies 6t(-t+1)=0$

$\implies t=0, 1$ second

We also know that rate of change of velocity is acceleration

Therefore,

$\boxed{a=\frac{dv}{dt}}$

$\implies a=\frac{d}{dt}(-6t^2+6t)$

$\implies a=-12t+6$

Acceleration at $t=0$ second

$a\Bigr|_{t=0}=0+6=6$ m/s²

Acceleration at $t=1$ second

$a\Bigr|_{t=1}=-12\times 1+6=-12+6=-6$m/s²

Hope this answer is helpful.