Question

The velocity of a particle moving in positive x-axis is according to relation $V = \sqrt {4x}$ .If at $t = 0$ particle is at $x = 0,$ then the velocity and acceleration at $t = 2s$ are given as
A. $\left( 1 \right).$ $4m{s^{ - 1}},4m{s^{ - 2}}$
B. $\left( 2 \right).$ $2m{s^{ - 1}},2m{s^{ - 2}}$
C. $\left( 3 \right).$ $2m{s^{ - 1}},4m{s^{ - 2}}$
D. $\left( 4 \right).$ $4m{s^{ - 1}},2m{s^{ - 2}}$

Answer 1

In this problem we need to find velocity and acceleration. In the question the velocity in terms of displacement has been given by using this we will find acceleration by differentiating the given velocity with respect to displacement. The value of time is substituted in the given equation to obtain the acceleration value. Then the velocity will be calculated by using the formula $\left( {V = a \times t} \right)$

Complete step-by-step solution:
Given: $V = \sqrt {4x}$ ………. $\left( 1 \right)$ , time $t = 2s$ , $V = ?$ and $a = ?$
$V = 2\sqrt x$
$V = 2{\left( x \right)^{\dfrac{1}{2}}}$
On differentiating above equation with respect to $x$
$\dfrac{{dV}}{{dx}} = 2 \times \dfrac{1}{2} \times {x^{\dfrac{1}{2} - 1}}$ , Using the identity of differentiation $\left[ {\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}} \right]$
On simplifying above equation
$\dfrac{{dV}}{{dx}} = {x^{ - \dfrac{1}{2}}}$
$\dfrac{{dV}}{{dx}} = \dfrac{1}{{\sqrt x }}$ ………. $\left( 2 \right)$
We know that, $acceleration = \dfrac{{velocity}}{{time}}$
That is, $a = \dfrac{{dV}}{{dt}}$
On multiplying and dividing by $dx$ to the above equation we get
$a = \dfrac{{dV}}{{dx}} \times \dfrac{{dx}}{{dt}}$ ………. $\left( 3 \right)$
We known that, $\dfrac{{dx}}{{dt}} = V$
Therefore, acceleration $a = \dfrac{{dV}}{{dx}} \times V$ ………. $\left( 4 \right)$
Substituting equation $\left( 1 \right)$ and equation $\left( 2 \right)$ in equation $\left( 4 \right)$
$a = \sqrt {4x} \times \dfrac{1}{{\sqrt x }}$
$a = 2\sqrt x \times \dfrac{1}{{\sqrt x }}$
On simplifying above equation we get
$a = 2$
Therefore, acceleration $a = 2m{s^{ - 2}}$
We known that, $acceleration = \dfrac{{velocity}}{{time}}$
That is, $a = \dfrac{V}{t}$
$\Rightarrow V = a \times t$
Substituting the value of $a = 2m{s^{ - 2}}$ and $t = 2s$
$V = 2 \times 2$
$\Rightarrow V = 4$
Therefore velocity $V = 4m{s^{ - 1}}$
Hence, Option D is correct. That is $4m{s^{ - 1}},2m{s^{ - 2}}$

Note: In mechanics the acceleration is defined as the rate of change of the velocity with respect to time. Whereas velocity is defined as the rate of change of position with respect to the frame of reference and it is also a function of time. Both acceleration and velocity is a vector quantity which has both magnitude and direction.