Question

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 1

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  

Answer 2

Answer:

v= 3t²-6t

at rest v=0

3t²-6t=0 (solve the quadratic)

t = 0 , t = 2

differentiate wrt time

dv/dt = 3t²-6t

= {6t-6}

put t=0 once then a=-6

put t=2 then a=6

hence answer is 6 and -6