Question

A particle moves so that\[S = 6 + 48t - {t^3}\]. Then after the particle moves, at what distance does the direction of motion reverse?
A. 63
B. 104
C. 134
D. 288

Answer 1

Hint: Before we start addressing the problem, we need to know about the displacement of a particle. It is defined as the change in the position of an object. Since it is a vector quantity it has both direction and magnitude.

Formula Used:
To find the velocity of a particle the formula is,
\[v = \dfrac{{ds}}{{dt}}\]
Where, \[ds\] is change in distance and \[dt\] is change in time.

Complete step by step solution:
Suppose a particle is moving from point A to point B at a velocity of v. When the particle reaches point B and reverses back to point A the velocity of the particle changes from v to zero and then back to v in the opposite direction.
Therefore, by the formula we have,
\[v = \dfrac{{dS}}{{dt}} = 0\]
\[\Rightarrow S = 6 + 48t - {t^3}\]………. (1)

Differentiate s with respect to the time,
\[\dfrac{{dS}}{{dt}} = 0 + 48 - 3{t^2}\]
\[\Rightarrow 48 - 3{t^2} = 0\]
\[\Rightarrow 3{t^2} = 48\]
\[\Rightarrow {t^2} = 16\]
\[\Rightarrow t = \pm 4\]
\[ \Rightarrow t = 4\]

Now, substitute the value of t in equation (1) we get,
\[S = 6 + 48\left( 4 \right) - {\left( 4 \right)^3}\]
\[\Rightarrow S = 6 + 192 - 64\]
\[ \therefore S = 134m\]
Therefore, the distance at which the particle reverses its direction is 134m.

Hence, Option C is the correct answer

Note: Basically the displacement tells the direct length between any two points when measured along the minimum path between them. Some of the examples of displacement are, a teacher walking across the blackboard, a jogger on a jogging track, etc. The displacement of an object value can be positive, negative, and even zero, but the distance can have only positive values and cannot be negative.