When displacement of a body is directly proportional to time?

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Answer

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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=\dfrac{dx}{dt}$$a=\dfrac{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=\dfrac{dx}{dt}=\dfrac{d}{dt}\left( k{{t}^{2}} \right)$$\Rightarrowv=\dfrac{dx}{dt}=\dfrac{d}{dt}\left( k{{t}^{2}} \right)$$\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=\dfrac{dv}{dt}=\dfrac{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 $x=ut+\dfrac{1}{2}a{{t}^{2}}$, 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 $x=\dfrac{1}{2}a{{t}^{2}}$.$\Rightarrow x\propto {{t}^{2}}$.Hence, the body is travelling with uniform acceleration.