A particle is moving so that its displacement s is given as s = t3- 6t2+ 3t + 4 meter. Its velocity at the instant when its acceleration is zero will be-a)3 m/sb)-12 m/sc)42 m/sd)-9 m/sCorrect answer is option 'D'. Can you explain this answer?

Question

Answer

For the given particle, v = 3t² - 12t + 3
And its acceleration, a = 6t - 12
Thus we get a = 0
At t = 2 sec,
Hence v at t = 2 sec
V = 3 x 4 - 12 x 2 + 3
= -9m/s

Answer

Given:
Displacement of the particle, s = t^3 - 6t^2 + 3t + 4 meter

Acceleration is given by the second derivative of displacement with respect to time:
a = d^2s/dt^2

To find the velocity at the instant when acceleration is zero, we need to find the time instant when the acceleration is zero.

Differentiating the displacement equation with respect to time, we get the velocity equation:
v = ds/dt

Differentiating the velocity equation with respect to time, we get the acceleration equation:
a = dv/dt = d^2s/dt^2

Differentiating the displacement equation twice, we get the acceleration equation:
a = d^2s/dt^2 = 6t - 12

Setting the acceleration equation equal to zero to find the time instant when the acceleration is zero:
6t - 12 = 0
6t = 12
t = 2 seconds

Now, substitute the value of t into the displacement equation to find the displacement at t = 2 seconds:
s = (2^3) - 6(2^2) + 3(2) + 4
s = 8 - 24 + 6 + 4
s = -6 meters

Substitute the value of t into the velocity equation to find the velocity at t = 2 seconds:
v = ds/dt = d/dt(t^3 - 6t^2 + 3t + 4)
v = 3t^2 - 12t + 3
v = 3(2^2) - 12(2) + 3
v = 12 - 24 + 3
v = -9 m/s

Therefore, the velocity at the instant when the acceleration is zero is -9 m/s. Hence, the correct answer is option 'D'.

Answer

C.-9 s=t^3-6t^2+3t+4 ---------1 v=3t^2-12t+3------------2 a=6t-12 (a=0)---------3 12=6t 3=t sub t in eq 2 v=3 (2)^2-12 (2)+3 =(12-24)+3 =-9

Answer

By differentiation we can get the answer

Answer

For the given particle, v = 3t² - 12t + 3
And its acceleration, a = 6t - 12
Thus we get a = 0
At t = 2 sec,
Hence v at t = 2 sec
V = 3 x 4 - 12 x 2 + 3
= -9m/s

Answer

For the given particle, v = 3t² - 12t + 3
And its acceleration, a = 6t - 12
Thus we get a = 0
At t = 2 sec,
Hence v at t = 2 sec
V = 3 x 4 - 12 x 2 + 3
= -9m/s

Answer

3 m/s

Answer

For the given particle, v = 3t² - 12t + 3
And its acceleration, a = 6t - 12
Thus we get a = 0
At t = 2 sec,
Hence v at t = 2 sec
V = 3 x 4 - 12 x 2 + 3
= -9m/s

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