Question

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to ________ .
A) ${t^{1/2}}$
B) $t$
C) ${t^{3/2}}$
D) ${t^2}$

Answer 1

Hint: Power refers to the work done by a body in a given time. The work done by the body is essentially the product of the applied force and the corresponding displacement of the body. Using these relations we can determine the relationship between the displacement of the body and the time using the method of dimensions.

Formula Used:
1) Power is given by, $P = \dfrac{W}{t}$ where $W$ is the work done and $t$ is the time taken to complete the work.
2) The velocity of a body is given by, $v = \dfrac{d}{t}$ where $d$ is the displacement of the body and $t$ is the time taken for the corresponding displacement.
3) The work done by an applied force $F$ is given by, $W = Fd$ where $d$ is the displacement of the body due to the applied force.

Complete step by step answer:
Step 1: Using the relations of the power and the work done by the body obtain a relationship between the displacement $d$ and time $t$ of the body.
The work done by an applied force $F$ on the given body can be expressed as $W = Fd$ ------(1)
The power of the source is mentioned to be constant.
i.e., $P = \dfrac{W}{t} = {\text{constant}}$ -------- (2)
Substituting equation (1) in (2) we get, $\dfrac{{Fd}}{t} = {\text{constant}}$ -------- (3)
Step 2: Using the method of dimensions, obtain the required relationship.
The dimension of force will be $F \to \left[ {ML{T^{ - 2}}} \right]$ as the unit of force is ${\text{N}}$ or ${\text{kgm}}{{\text{s}}^{ - 2}}$.
The dimension of displacement will be $d \to \left[ L \right]$ as the unit of displacement is meters.
And the dimension of time is $t \to \left[ T \right]$ as the unit of time is seconds.
Now rewrite equation (3) in terms of the dimensions of the force, displacement and time.
i.e., $\dfrac{{\left[ {ML{T^{ - 2}}} \right]\left[ L \right]}}{{\left[ T \right]}} = {\text{constant}}$
Simplifying we get, $M{L^2}{T^{ - 3}} = {\text{constant}}$ or ${L^2}{T^{ - 3}} = {\text{constant}}$
$ \Rightarrow {L^2} \propto {T^3}$
Then on taking the square root, we have $L \propto {T^{3/2}}$
Since length is the dimension of displacement, we have $d \propto {t^{3/2}}$ .
Thus the correct option is C.

Note: Alternate method
We have the power of the source as $P = \dfrac{W}{t}$ where the work done by the body is expressed as $W = Fd$ .
Then we express the power of the source as $P = \dfrac{{Fd}}{t}$
Now as the velocity of the body is given by, $v = \dfrac{d}{t}$ and the force is given by, $F = ma$ where $m$ is the mass and $a$ is the acceleration of the body, the power can be expressed as $P = mav$ .
Finally, we substitute $v = at$ to obtain $P = m{a^2}t$
$ \Rightarrow a = \sqrt {\dfrac{P}{{mt}}} $ ------ (A)
Now Newton’s first equation of motion gives the displacement of the body as
$d = ut + \dfrac{1}{2}a{t^2}$ ---------- (B).
We assume that the body was initially at rest, i.e., $u = 0$ .
So equation (B) becomes $d = \dfrac{1}{2}a{t^2}$ ------ (C)
Substituting equation (A) in (C) we get, $d = \dfrac{1}{2}\left( {\sqrt {\dfrac{P}{{mt}}} } \right){t^2}$
On simplifying we get, $d = \dfrac{1}{2}\sqrt {\dfrac{P}{m}} \times {t^{\left[ {2 - \left( {\dfrac{1}{2}} \right)} \right]}} = \dfrac{1}{2}\sqrt {\dfrac{P}{m}} \left( {{t^{3/2}}} \right)$
$ \Rightarrow d \propto {t^{3/2}}$
So the correct option is C.