The velocity of a particle moving on straight line is given as v = 3t^

1.	The velocity of a particle moving on straight line is given as v = 3t^2 - 6t. The acceleration of particle when body is at rest-(A) Zero(B) 6 m/s^2(C)-6 m/s^2(D) Both (B) and (C)(Answer is D. Please mention reason or steps for solution.)
Answer» Answer: EXPLANATION: GIVEN The VELOCITY of a particle moving on straight line is given as v = 3t^2 - 6t. The acceleration of particle when BODY is at rest- Now given 3t^2 – 6 t Velocity is zero when body is at rest. Now 3t^2 – 6 t = 0 3 t(t – 2) = 0 t = 0, t = 2 v = 3 t^2 – 6 t Differentiate with respect to t we get dv/dt = d/dt(3 t^2 – 6 t) = 6t - 6 So acceleration for t = 0 will be = 6(0) – 6 = - 6 So acceleration at t = 2 will be 6(2) – 6 = 6

The velocity of a particle moving on straight line is given as v = 3t^2 - 6t. The acceleration of particle when body is at rest-(A) Zero(B) 6 m/s^2(C)-6 m/s^2(D) Both (B) and (C)(Answer is D. Please mention reason or steps for solution.)

Answer»

Answer:

GIVEN The VELOCITY of a particle moving on straight line is given as v = 3t^2 - 6t. The acceleration of particle when BODY is at rest-

Now given 3t^2 – 6 t

Velocity is zero when body is at rest.

Now 3t^2 – 6 t = 0

3 t(t – 2) = 0

t = 0, t = 2

v = 3 t^2 – 6 t

Differentiate with respect to t we get

dv/dt = d/dt(3 t^2 – 6 t)

= 6t - 6

So acceleration for t = 0 will be = 6(0) – 6 = - 6

So acceleration at t = 2 will be 6(2) – 6 = 6