Question 105: Motion in 1D

Question

A particle is moving in one dimension (along $x$ axis) under the action of a variable force. It's initial position was $16 \mathrm{~m}$ right of origin. The variation of its position $(x)$ with time $(t)$ is given as $x=-3 t^3+18 t^2+16 t$, where $x$ is in $\mathrm{m}$ and $\mathrm{t}$ is in $\mathrm{s}$. The velocity of the particle when its acceleration becomes zero is _________ $\mathrm{m} / \mathrm{s}$.

Accepted Answer

Position (x):
The particle's location on the x-axis at a given time.
Velocity (v):
The rate of change of position with respect to time
( $v = \frac{dx}{dt}$ ).
Acceleration (a):
The rate of change of velocity with respect to time
( $a = \frac{dv}{dt}$ ).
The particle's position is given by:
$$x(t) = -3t^3 + 18t^2 + 16t$$
We need to find the velocity when the acceleration is zero.
Find the Velocity (v) and Acceleration (a) Functions:
Velocity:
$$v(t) = \frac{dx}{dt} = -9t^2 + 36t + 16$$
Acceleration:
$$a(t) = \frac{dv}{dt} = -18t + 36$$
Find the Time (t) When Acceleration is Zero:
$$a(t) = 0$$
$$-18t + 36 = 0$$
$$t = 2 	ext{ seconds}$$
Calculate the Velocity at t = 2 seconds:
$$v(2) = -9(2)^2 + 36(2) + 16$$
$$v(2) = 52 	ext{ m/s}$$
Answer:
The velocity of the particle when its acceleration becomes zero is 52 m/s.

Motion in 1D

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