Question

If the displacement of an object is proportional to the square of time, then the object moves with

• A. uniform velocity
• B. uniform acceleration
• C. increasing acceleration
• D. decreasing acceleration

Answer 1

The correct option is B

uniform acceleration

Step1: Given data

It is given that the displacement of a body is proportional to the square of time.

Step2: Formula used

$$\begin{gathered} v=\frac{dx}{dt}[v=velocity,\frac{dx}{dt}=changeindisplacementovertime] \\ a=\frac{dv}{dt}[a=acceleration,\frac{dv}{dt}=changeinvelocityovertime] \end{gathered}$$

Step3: Calculating acceleration

Let us assume that $x$ is the displacement of the body and t is time.

According to the given data, $x\propto t^2.............\left(i\right)$

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 $x=k{t}^2\dots ..\left(ii\right).$

We can see that the options are stating 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 $t$.

$$\begin{gathered} v=\frac{dx}{dt}=\frac{d}{dt}(k{t}^{2)} \\ v=2kt.............(iii) \end{gathered}$$

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 $t$.

$$\begin{gathered} a=\frac{dv}{dt}=\frac{d}{dt}\left(2kt\right) \\ a=2k \end{gathered}$$

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

Hence, option B is the correct answer.