Question

The displacement $x$ of a particle moving in one dimension under the action of a constant force is related to time $t$ by the equation $t-3=\sqrt{x}$ , where $x$ is in meter and $t$ is in second. Find the displacement of the particle when its velocity is zero.

Answer 1

- Hint- In order to find the displacement of the particle first we will find the velocity by differentiating the given equation with respect to $t$ . In this way we will calculate the time, and by putting this time in the given equation we will get the displacement.

Complete step-by-step solution -

Given equation is $t-3=\sqrt{x}$ --- (1)
Let us convert our equation in simplest form, so after taking square to both the side we have,
$$\begin{gathered} \because t-3=\sqrt{x} \\ \Rightarrow {\left(t-3\right)}^2={\left(\sqrt{x}\right)}^2 \end{gathered}$$
Now let us simplify the term by using the algebraic formula for the square of sum of terms.
$$\begin{gathered} \Rightarrow {\left(t-3\right)}^2={\left(\sqrt{x}\right)}^2 \\ \Rightarrow t^2+3^2-2\times 3\times t=x\left[\because {\left(a-b\right)}^2=a^2+b^2-2\times a\times b\right] \\ \Rightarrow t^2+9-6t=x \\ \Rightarrow x=t^2-6t+9 \end{gathered}$$
As we know that the velocity is the time rate of change of displacement.
$$\begin{gathered} \Rightarrow {\displaystyle \frac{d}{dt}}\left[x\right]={\displaystyle \frac{d}{dt}}\left[t^2-6t+9\right] \\ \Rightarrow {\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d}{dt}}\left[t^2\right]+{\displaystyle \frac{d}{dt}}\left[-6t\right]+{\displaystyle \frac{d}{dt}}\left[9\right] \\ \Rightarrow v={\displaystyle \frac{dx}{dt}}=2t+\left(-6\right)+0 \\ \Rightarrow v=2t-6 \end{gathered}$$
Given the problem, we need to find out the displacement when the velocity is zero.
So let us equate the equation of velocity to zero to find the time $t$
$$\begin{gathered} \Rightarrow v=2t-6=0 \\ \Rightarrow 2t=6 \\ \Rightarrow t={\displaystyle \frac{6}{2}} \\ \Rightarrow t=3\sec \end{gathered}$$
Therefore, the value of time t is 3 sec.
Now we will substitute value of time in given equation (1), we obtain
$$\begin{gathered} \because t-3=\sqrt{x} \\ \Rightarrow \left(3\right)-3=\sqrt{x}\left[\because \text{ found the value of t = 3 sec}\right] \\ \Rightarrow 0=\sqrt{x} \\ \Rightarrow {\left(0\right)}^2={\left(\sqrt{x}\right)}^2 \\ \Rightarrow x=0 \end{gathered}$$
Hence, the displacement of the particle is 0 when its velocity is zero.

Note- In order to solve such problems students must remember that by single differentiation of displacement equation with respect to time is velocity and double differentiation of displacement equation with respect to time is acceleration. A displacement is a vector whose length is the shortest distance from the initial to the final position of a point P undergoing motion. The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time.