Question

If the displacement of a body is proportional to the square of time, then the body is moving with
A. Uniform acceleration
B. Increasing acceleration
C. Decreasing acceleration
D. Uniform velocity

Answer 1

Hint:Write the equation for the displacement of the body as a function of time. Then differentiate the displacement with respect to time and find the velocity. After this, differentiate the velocity with respect to time to find the acceleration.

Formula used:
$v={\displaystyle \frac{dx}{dt}}$
$a={\displaystyle \frac{dv}{dt}}$

Complete step by step answer:
It is given that the displacement of a body is proportional to the square of time.

Therefore, we can write that $x\propto t^2$ …. (i),
where x is the displacement of the body and t is time.

This means that if the time is increased by factor ‘n’ then the displacement of the particle will increase by a factor of $n^2$.

By adding a proportionality constant k to (i), we can write an equation for x as $x=k{t}^2$ ….. (ii).

We can see that the options are stating about the velocity and the acceleration of a body.
Velocity (v) of a body is the first derivative of displacement with respect to time.

Therefore, differentiate (ii) with respect to time t.
$\Rightarrow v={\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d}{dt}}\left(k{t}^2\right)$$\Rightarrow v={\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d}{dt}}\left(k{t}^2\right)$
$\Rightarrow v=2kt$ …. (iii)

Now, we can see that the velocity of the given body is directly proportional to t.
Acceleration (a) of a body is the first derivative of its velocity with respect to time t.

Therefore, differentiate (iii) with respect to time t.
$\Rightarrow a={\displaystyle \frac{dv}{dt}}={\displaystyle \frac{d}{dt}}\left(2kt\right)$
$\Rightarrow a=2k$.

We know that k is a constant. Therefore, the acceleration of the body is constant.
Constant acceleration is also called uniform acceleration.

Hence, the correct option is A.

Note:If you are well known with the kinematic equations for uniform acceleration, then this would be a very easy problem.

One of the kinematic equations say that $x=ut+{\displaystyle \frac{1}{2}}a{t}^2$, where u is the initial velocity of the body.

If we put a condition that the body was at rest initially ( $u=0$), then the displacement of the body is equal to $x={\displaystyle \frac{1}{2}}a{t}^2$.
$\Rightarrow x\propto t^2$.

Hence, the body is travelling with uniform acceleration.