Question

The velocity $v$ and displacement r of a body are related as \[{v^2} = kr,\] where k is a constant. What will be the velocity after 1 second? (Given that the displacement is zero at t = 0).
A. \[\sqrt {kr} \]
B. \[k{r^{\dfrac{3}{2}}}\]
C. \[\dfrac{k}{2}{r^0}\]
D. Data is not sufficient

Answer 1

Hint: The equations of motion can help us to relate the movement of a body in terms of position, velocity, acceleration in the given period of time. The motion of an object can be in different paths, but we have a body moving in one dimension, which can be used with the equations of motion.

Formula used :
The velocity of the body can be given as,
\[v = u + at\]
Where, v - final velocity, u – initial velocity, a – acceleration and t – time.

Complete step by step solution:
In physics, there are many types of motion, which can be described using various types of equations. Here, we discuss the motion of a body in one dimension, hence we can use the equations of motion that relate the displacement, distance, velocity, acceleration and time. In this question we have given the velocity and displacement, with which we need to find the final velocity of the body after 1 second. So, we can use the formula for it as,
\[v = u + at\]

In order to use this formula, we need acceleration, we know that differentiating the velocity gives acceleration hence we can proceed with it. Given,
\[{v^2} = kr\]
We can write it as,
\[v = \sqrt {kr} \]
Differentiating the above equation,
\[\dfrac{{dv}}{{dt}} = \sqrt k \left( {\dfrac{1}{2}} \right){r^{ - \dfrac{1}{2}}}\dfrac{{dr}}{{dt}}\]

From this, we can write the acceleration as,
\[a = \sqrt k \left( {\dfrac{1}{2}} \right){r^{ - \dfrac{1}{2}}}\dfrac{{dr}}{{dt}}\]
\[\Rightarrow a = \sqrt k \left( {\dfrac{1}{2}} \right){r^{ - \dfrac{1}{2}}}.v\]
\[\Rightarrow a = \sqrt k \left( {\dfrac{1}{2}} \right){r^{ - \dfrac{1}{2}}}.\sqrt {kr} \]
\[\Rightarrow a = \sqrt k \left( {\dfrac{1}{2}} \right){r^{ - \dfrac{1}{2}}}.\sqrt k {r^{\dfrac{1}{2}}}\]
\[\Rightarrow a = \left( {\dfrac{k}{2}} \right){r^0}\]

Now using the equation of motion,
\[v = u + at\]
Given, u = 0 m\s, t = 1 s
\[a = \left( {\dfrac{k}{2}} \right){r^0}\]
To find, v = ?
\[v = 0 + \dfrac{k}{2}{r^0}\left( 1 \right)\]
\[\therefore v = \dfrac{k}{2}{r^0}\]

So, the correct answer is option C.

Note : When you differentiate the velocity, you are differentiating the velocity with respect to time, so constant \[\sqrt k \] remains as it is and only the variable which is the displacement is differentiated with respect to time. If displacement is differentiated once then, velocity is found and if it is differentiated twice, then acceleration is found.