Question

The equation of motion of a particle is given by $s = 2{t^3} - 9{t^2} + 12t + 1$ , where $s$ and $t$ are measured in cm and s. The time when the particle stops momentarily is
A. $1$ sec
B. $2$ sec
C. $1,2$ sec
D. None of these

Answer 1

Hint: The position of a particle at any time is given to us, we have to find the time when the particle stops momentarily. Speed is defined as the rate at which an object's location changes in any direction. The distance traveled divided by the time it took to travel that distance is how fast something is moving, that is, $speed = \dfrac{s}{t}$ , when the value of distance and time are very small then we use the form $speed = \dfrac{{ds}}{{dt}}$ , that means, we differentiate the distance with respect to time. So, we will find the speed of the given particle using the above information and thus find the time.

Complete step by step solution:
We are given the equation of motion of particles as $s = 2{t^3} - 9{t^2} + 12t + 1$. We have to find the value of $t$ when the particle stops momentarily. We know that $\dfrac{{ds}}{{dt}}$ gives the speed of the object. So, the speed of this particle will be:
\[\dfrac{{ds}}{{dt}} = \dfrac{{d(2{t^3} - 9{t^2} + 12t + 1)}}{{dt}} \\
\Rightarrow \dfrac{{ds}}{{dt}} = \dfrac{{d(2{t^3})}}{{dt}} - \dfrac{{d(9{t^2})}}{{dt}} + \dfrac{{d(12t)}}{{dt}} + \dfrac{{d(1)}}{{dt}} \\ \]
We know that $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ and $\dfrac{{dk}}{{dx}} = 0$ , so:
\[\dfrac{{ds}}{{dt}} = 2(3{t^2}) - 9(2t) + 12 + 0 \\
\Rightarrow \dfrac{{ds}}{{dt}} = 6{t^2} - 18t + 12 \\ \]
When a particle is not moving momentarily, its speed is zero, so $\dfrac{{ds}}{{dt}} = 0$
$\Rightarrow 6{t^2} - 18t + 12 = 0 \\
\Rightarrow {t^2} - 3t + 2 = 0 \\ $
Solving the above quadratic equation using factorization, we get:
${t^2} - t - 2t + 2 = 0 \\
\Rightarrow t(t - 1) - 2(t - 1) = 0 \\
\Rightarrow (t - 2)(t - 1) = 0 \\
\therefore t = 1,2 $

The correct option is option C.

Note: When the object is stopped momentarily, the distance covered by it in time $dt$ is zero, that is, the position remains constant as the time passes. We know that the derivative of a constant is zero, so the speed of the particle is zero when it stops momentarily.