Question

A body is moving along a straight line by a machine delivering constant power. The distance covered by the body in time t is proportional to,
$\begin{aligned} & a){{t}^{1/2}} \\\ & b){{t}^{3/4}} \\\ & c){{t}^{3/2}} \\\ & d){{t}^{2}} \\\ \end{aligned}$

Answer 1

It is given in the question that the body moves due to the supply of constant power from the machine. Hence we first obtain the expression for the power supplied to the body at time t. Further using Newton's second kinematic equation we will express the expression for power in terms of displacement of the body with time. Further from the obtained expression we will accordingly determine the variation of distance covered with time.

Formula used:
$S=Ut+\dfrac{1}{2}a{{t}^{2}}$
$P=\dfrac{W}{t}=\dfrac{FS}{t}=\dfrac{maS}{t}$

Complete step by step answer:
Let us say the above body mentioned is initially at rest i.e. its initial velocity (U) is zero. The body is supplied with constant power, hence it will move with uniform acceleration(a). Hence the distance covered by the body in time t using Newton’s second kinematic equation we get,
$\begin{aligned} & S=Ut+\dfrac{1}{2}a{{t}^{2}}\text{, }\because \text{U=0} \\\ & \Rightarrow S=(0)t+\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow S=\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow a=\dfrac{2S}{{{t}^{2}}} \\\ \end{aligned}$
Let us say the energy supplied by the machine in time t be (W). The power supplied is defined as the rate at which energy is given to the body. Since the power is constant, mathematically the power (P)supplied is equal to,
$P=\dfrac{W}{t}$
The work done or the energy supplied by the machine is equal to the product of force by which the times the distance covered by the body in time t. And Newton's second law force is given as the product of mass ‘m’ times the acceleration. Hence the above equation of power becomes,
$\begin{aligned} & P=\dfrac{W}{t} \\\ & \Rightarrow P=\dfrac{FS}{t} \\\ & \Rightarrow P=\dfrac{maS}{t} \\\ \end{aligned}$
Substituting for acceleration in the above equation we get,
$\begin{aligned} & P=a\dfrac{mS}{t} \\\ & \Rightarrow P=\dfrac{2S}{{{t}^{2}}}\dfrac{mS}{t}=\dfrac{2m{{S}^{2}}}{{{t}^{3}}} \\\ & \Rightarrow {{S}^{2}}=\dfrac{P{{t}^{3}}}{2m} \\\ & \Rightarrow S=\sqrt{\left( \dfrac{P}{2m} \right)}{{t}^{3/2}} \\\ & \therefore S\propto {{t}^{3/2}} \\\ \end{aligned}$

Therefore the correct answer of the above question is option C.

Note:
It is to be noted that initial velocity we have considered to be zero as we have assumed that the body is initially solely moving due to the machine. Also we were in a position to use the expression for power as the power supplied by the machine is constant. This basically means for every unit time the energy supplied is the same throughout.