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

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Last updated date: 23rd Feb 2024
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IVSAT 2024
Answer
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Hint: In order to find the solution of the given question we need to know the equations of motion. After applying the equations of motion, we can finally conclude with the correct solution for the given question. First of all we need to differentiate the position with respect to time to get the velocity. After that we need to differentiate the velocity in order to get the required value of the acceleration.

Complete step by step solution:
The position of the particle in the question is given as, $x = ({t^3} - 6{t^2} - 15t + 40)m$
We know that velocity, $v = \dfrac{{dx}}{{dt}} = \dfrac{{d({t^3} - 6{t^2} - 15t + 40)}}{{dt}}$
$\therefore v = 3{t^2} - 12t - 15m{s^{ - 1}}$
Also, the acceleration of a body is given by, $a = \dfrac{{dv}}{{dt}} = \dfrac{{d(3{t^2} - 12t - 15)}}{{dt}}$
$\therefore a = 6t - 12m{s^{ - 2}}$
a) We need to find the time at which the velocity is zero.
Therefore, we can write the equation as,
$\Rightarrow v = 3{t^2} - 12t - 15 = 0$
$ \Rightarrow 3{t^2} - 15t + 3t - 15 = 0$
$ \Rightarrow 3t(t - 5) + 3(t - 5) = 0$
$ \Rightarrow (3t + 3)(t - 5) = 0$
$\therefore t = - 1$ or $t = 5$
Since, the time cannot be negative, therefore, $t = 5s$

b) Now we need to find the position at $t = 5s$
$\Rightarrow x = ({t^3} - 6{t^2} - 15t + 40)m$
$ \Rightarrow x = {5^3} - 6 \times {5^2} - 15 \times 5 + 40$
$ \Rightarrow x = 125 - 150 - 75 + 40 = - 60m$
Similarly, position at $t = 0s$
$\Rightarrow x = ({t^3} - 6{t^2} - 15t + 40)m$
$ \Rightarrow x = {0^3} - 6 \times {0^2} - 15 \times 0 + 40 = 40m$
Now, we need to find the displacement at $t = 5s$ and $t = 0s$
$\Rightarrow s = {x_5} - {x_0} = - 60 - 40 = - 100m$

c) Now, we need to find the acceleration at $t = 5s$
$\Rightarrow a = 6t - 12 = 6 \times 5 - 12$
$\therefore a = 30 - 12 = 18m{s^{ - 2}}$
Therefore, the required acceleration is $18m{s^{ - 2}}$.

Note: We define velocity of an object as the rate of change of its position with respect to reference point. It is the function of time. We define acceleration as the rate of change of velocity of the object with respect to the frame of time. It is a vector quantity which means it has magnitude and direction. When an object accelerates it means the velocity keeps changing. When velocity is zero it means that there is no acceleration.