Question

A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as x = –3t3 + 18t2 + 16t, where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m/s.

Answer 1

We are given the position equation:

$x = -3t^3 + 18t^2 + 16t$

To find the velocity, we differentiate $x(t)$ with respect to $t$:

$v = \frac{dx}{dt} = -9t^2 + 36t + 16$

Next, we differentiate again to find the acceleration:

$a = \frac{dv}{dt} = -18t + 36$

Now, set the acceleration equal to zero to find the time when acceleration becomes zero:

$-18t + 36 = 0$

Solving for $t$:

$t = 2 \, \text{s}$

Substitute $t = 2$ into the velocity equation:

$v = -9(2)^2 + 36(2) + 16 = -36 + 72 + 16 = 52 \, \text{m/s}$

Thus, the velocity of the particle when its acceleration becomes zero is 52 m/s.