Question

Motion of a particle is given by equation s=(3t^3+7t^2+14t+8)m. the value of acceleration of the particle at t=1 sec is

Answer 1

Answer:

The acceleration of the particle at t = 1 sec is 32 m/s².

Explanation:

Motion of a particle is given by equation,

$s=(3t^3+7t^2+14t+8)\ m$.....(I)

We need calculate the acceleration of the particle

The velocity is the first derivative of the position.

The velocity is

On differentiating of equation (I) with respect to time

$v =\dfrac{ds}{dt}=9t^2+14t+14$

The acceleration is the rate of change of the velocity.

On differentiating again with respect to time

$a =\dfrac{dv}{dt}=18t+14$

The acceleration of the particle at t = 1 sec is

$a = 18\times1+14$

$a = 32\ m/s^2$

Hence, The acceleration of the particle at t = 1 sec is 32 m/s².

Answer 2

Answer:

___________32m/s^2 ___________