Question

A particle moves along a straight line OX. At a time t (in seconds) the distance x (in meters) of the particle from 0 is given by: $$x=40+12t-t^3$$. How long would the particles travel before coming to rest?
A) 24m
B) 40m
C) 56m
D) 16m

Answer 1

Hint: The distance is defined as the length of the path between the initial position and the final position of a moving object in the given interval of time. It is a scalar quantity. The change in the position of an object is called displacement. Displacement is a vector quantity and has both magnitude and direction.

Complete step by step solution:
Given equation $$x=40+12t-t^3$$.....................(1)
Let say that the particle is at a distance x from O at a time, t=0
Thus substituting the value of t=0 in the given equation, $$x=40+12t-t^3$$, we get,
$$x=40+12\left(0\right)-{\left(0\right)}^3=40m$$
It is given that the particle will come to rest which means the velocity of the particle becomes zero after traveling to a certain displacement and let us consider this time as t.
Differentiating the given equation 1 to time, we get,
$\Rightarrow v=12-3t^2$.................(2)
When the particle comes to rest at time=t, velocity=0
Substituting the value of v=0 in equation 2, we get,
$\Rightarrow 12-3t^2=0$
$\Rightarrow 3t^2=12$
$\Rightarrow t^2=4$
$\Rightarrow t=2s$
Substituting the value of t in equation 1, we get
$$\Rightarrow x=40+12\left(2\right)-{\left(2\right)}^3$$
$$\Rightarrow x=56m$$
Thus, it is clear that the particle had started the journey when it was at a distance of 40m from point O and came to rest at 56m from point O.
Thus, the particle had traveled a distance of $56m-40m=16m$.
Thus, the particle had traveled a distance of 16m before coming to rest.

Hence the correct option is D.

Note: 1. The velocity is defined as the rate of change of the position of the object to time. Velocity is also a vector quantity and has both direction and magnitude.
2. The distance and the displacement have the same SI unit, meter.