Question

The displacement $\left( x \right)$ and time $\left( t \right)$ for a particle are related $t = \sqrt x + 3$. What is the work done by the particle in the first six-second of its motion?
(A) $3J$
(B) $4J$
(C) $Zero$
(D) $5J$

Answer 1

Hint The rate of change of displacement is called the velocity of the particle. The rate of change of a quantity to time can be found by differentiating the quantity with time. So, first, differentiate the displacement equation to find the velocity. Then find the work done by the particle.
Formula used
Work done = change in Kinetic energy $K.E$ $= K.{E_{final}} - K.{E_{initial}}$

Complete Step-by-step solution
Given equation is $t = \sqrt x + 3$ ;
Let us convert our equation in the simplest form,
$t - 3 = \sqrt x$
Taking square to both the side we get,
$\Rightarrow {\left( {t - 3} \right)^2} = {\left( {\sqrt x } \right)^2}$
Shorten the term by using the algebraic formula for the square of the sum of terms.
$\Rightarrow {t^2} + {3^2} - 2 \times 3 \times t = x$
$\Rightarrow x = {t^2} - 6t + 9$
As we know that the velocity is the rate of change of displacement.
$\Rightarrow \dfrac{d}{{dt}}\left[ x \right] = \dfrac{d}{{dt}}\left[ {{t^2} - 6t + 9} \right]$
$\Rightarrow v = \dfrac{{dx}}{{dt}} = 2t + \left( { - 6} \right) + 0$
$\Rightarrow v = 2t - 6$
Given the problem, we need to find out the work done in the first six-second of its motion.
At $t = 0$ , $v = - 6m{s^{ - 1}}$
At $t = 6s$, $v = 6m{s^{ - 1}}$
As we know the Formula for calculating work done is
Work done = change in Kinetic energy $K.E$ $= K.{E_{final}} - K.{E_{initial}}$
$\Rightarrow \dfrac{1}{2}m{v_{final}}^2 - \dfrac{1}{2}m{v_{initial}}^2$
On relating the values in the above equation we get,
$\Rightarrow \dfrac{1}{2}m{\left( 6 \right)^2} - \dfrac{1}{2}m{\left( { - 6} \right)^2}$
$\Rightarrow 0J$

Hence, the correct option is (C) Zero.

Additional Information A displacement is the shortest distance from the initial to the final position. It is a vector quantity. The work-energy theorem states that the work done by the sum of all forces acting on a body equals the change in kinetic energy of the body.

Note Students must remember that single differentiation of displacement equation with respect to time is velocity and double differentiation of displacement equation with respect to time is Acceleration. Do not get confused between speed and velocity. Velocity is displacement per unit time and speed is distance per unit time.