Question

A particle is moving along a straight line and its position is given by the relation $x = \left( {{t^3} - 6{t^2} - 15t + 40} \right)m$ :
Find
(a) The time at which the velocity is zero.
(b)Position and displacement of the particle at that point
(c) Acceleration for particles at the line.

Answer 1

We know that the velocity of a body is nothing but its rate of change of displacement with respect to time. Similarly, we also know that acceleration of a body is nothing but the rate of change of velocity with respect to time. Thus, to find velocity of a body with its displacement known all we have to do is to differentiate it with respect to time. Similarly, to find acceleration with velocity of the body known we need to differentiate velocity with respect to time.

Formula used: We will be using the formula $v = \dfrac{{ds}}{{dt}}$ where $v$ is the velocity of the body, $s$ is the displacement of the body, and $t$ is the time taken by the body to travel the displacement $s$ . We will also be using the formula, $a = \dfrac{{dv}}{{dt}} = \dfrac{{{d^2}s}}{{d{t^2}}}$ where $a$ is the acceleration experienced by the body.

Complete Step by Step Answer:
We know that the body in motion can be defined by the distance or displacement travelled by the body, the rate of change of displacement with respect to time (the velocity of the body) and the rate of change of velocity with respect to time (the acceleration experienced by the body).
From the problem we can infer that the position of the particle is given by the relation, $x = \left( {{t^3} - 6{t^2} - 15t + 40} \right)m$ . We know that the velocity of the body is the rate of change of displacement.
$v = \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {{t^3} - 6{t^2} - 15t + 40} \right)$
Applying the basic rules of differentiation, we get,
$v = 3{t^2} - 12t - 15 + 0$
The equation can be further simplified by taking the common multiple 3 and dividing the equation by 3 on both sides.
$v = {t^2} - 4t - 5$
We are supposed to find the time at which the velocity of the particle would be zero. So, let us assume that $v = 0$ .
$0 = {t^2} - 4t - 5$
Solving by splitting the middle term we get,
$\left( {t - 5} \right)\left( {t + 1} \right) = 0$
Thus, the roots of the equation are $t = 5s$ or $t = - 1$ . Since time cannot be a negative integer the time at which the velocity of the body is zero will be $t = 5s$ .
The position of the particle at this time interval can be given by substituting the value of $t = 5s$ in $x$ .
$x = \left( {{{\left( 5 \right)}^3} - 6{{\left( 5 \right)}^2} - 15\left( 5 \right) + 40} \right)m$
$x = 125 - 150 - 75 + 40$
Solving for $x$ we get,
$x = - 60m$
Thus, the position of the particle is $60m$ in the negative x direction.
We are now required to find the acceleration of the particle at this time interval of $t = 5s$ . We know that acceleration can be given by, $a = \dfrac{{dv}}{{dt}}$ .
We found $v = {t^2} - 4t - 5$ to be the velocity of the particle. Differentiating $v$ to find acceleration.
$a = \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}\left( {3{t^2} - 12t - 15 + 0} \right)$
Applying the basic rules of differentiation, we get,
$a = 6t - 12$
We are required to find the acceleration at $t = 5$ . After substituting we get,
$a = 6\left( 5 \right) - 12$
$\Rightarrow a = 18m/{s^2}$

Thus, the acceleration of the body would be $a = 18m/{s^2}$ .

Note: Similarly, the value of velocity at any time can be calculated by substitution of the necessary variables in $x$ which is defined as a function in $t$. So we can calculate the value of the velocity.