Question

If velocity of the particle is given by $$v=\sqrt{x}$$ where x denotes the position of the particle and initially the particle was at $$x=4$$ then which of the following are correct?
(A) At $$t=2s$$ the position of the particle is $$x=9$$
(B) Particle's acceleration at $$t=2s$$ is $$1m/s^2$$
(C) Particle's acceleration is $${\displaystyle \frac{1}{2}}m/s^2$$ throughout the motion
(D) Particle will never go in negative direction from its starting position​

Answer 1

Hint: We can start to answer this question by noting down all the given data. Then once we have done that, we can move onto finding the relationship between the velocity and distance and from that we will be able to get the value of time. To find the acceleration of the particle, we can find the differential of the velocity with respect to time.
Formulas used:
The formula to find the velocity with respect to distance and time is given by, $${\displaystyle \frac{dx}{dt}}=v$$
The formula to find the acceleration with respect to velocity and time is given by, $${\displaystyle \frac{dv}{dt}}=a$$
The formula to find the acceleration with respect to distance and velocity is given by, $$a={\displaystyle \frac{dv}{dt}}\Rightarrow {\displaystyle \frac{dv}{dx}}\times {\displaystyle \frac{dx}{dt}}={\displaystyle \frac{dv}{dx}}v$$
Where v is the velocity of the particle and a is the acceleration of the particle

Complete answer:
Let us start by writing the given data
the velocity of the particle is given as $$v=\sqrt{x}$$ where x is the position of the particle
Using the formula of velocity, we the value of time,
$$v=\sqrt{x}$$
$${\displaystyle \frac{dx}{dt}}=v$$
Now we equate the two equations above and get
$${\displaystyle \frac{dx}{dt}}=\sqrt{x}$$
We rearrange this and get $${\displaystyle \frac{dx}{\sqrt{x}}}=dt$$
As we integrate this, we get the value of time as, $$\int {\displaystyle \frac{dx}{\sqrt{x}}}=\int dt$$
Solving this, we arrive at $$2\sqrt{x}=t+c$$
Where c is the constant of integration.
Now we apply the value of $$x=4$$ and get the value of c
That is, $$2\sqrt{4}=0+c\Rightarrow c=4$$
We can find the position of the particle at $$t=2s$$ as
$$2\sqrt{x}=t+c\Rightarrow 2\sqrt{x}=2+4$$
Taking position to on side, $$x={\displaystyle \frac{6^2}{4}}=9m$$
Now moving onto the acceleration of the particle, we have
$$a={\displaystyle \frac{dv}{dt}}\Rightarrow {\displaystyle \frac{dv}{dx}}\times {\displaystyle \frac{dx}{dt}}={\displaystyle \frac{dv}{dx}}v$$
We get $$a={\displaystyle \frac{dv}{dx}}v=\sqrt{x}\times {\displaystyle \frac{1}{2\sqrt{x}}}={\displaystyle \frac{1}{2}}m/s^2$$

This leads us to the answer that options (A) (C) and (D) are correct.

Note:
Velocity is defined as the ratio of the change in position of a particle to the change in time. Likewise, the acceleration of a particle can be defined as the ratio of the change in velocity to the change in time.