Question

The displacement x of a particle at the instant when its velocity is v is given by \[v = \sqrt {3x + 16} \]. Its acceleration and initial velocity are,
A. 1.5 units, 4 units
B. 3 units, 4 units
C. 16 units, 1.6 units
D. 16 units, 3 units

Answer 1

Hint:The velocity is the rate of change of position with respect to time. The initial velocity and the acceleration is the value of velocity and acceleration when the motion gets started to be recorded, i.e. at \[t = 0s\].

Formula used:
The third kinematic equation of motion,
\[{v^2} = {u^2} + 2as\]
Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

Complete step by step solution:
The given expression for the velocity at any position x is,
\[v = \sqrt {3x + 16} \]
Here, v is the velocity and x is the position of the body. We need to find the initial velocity and initial acceleration of the motion.
On squaring both the sides of the velocity expression, we get
\[{v^2} = 16 + 3x\]
On rearranging the terms, we get
\[{v^2} = {\left( 4 \right)^2} + 2 \times \left( {\dfrac{3}{2}} \right)x \ldots \left( i \right)\]
The third kinetic equation of motion is given as,
\[{v^2} = {u^2} + 2ax \ldots \left( {ii} \right)\]
When we compare both the equations, we get
\[u = 4\,units\]
\[\therefore a = \dfrac{3}{2}\,units = 1.5\,units\]
As u is initial velocity and a is the acceleration, so the initial velocity is 4 units and the initial acceleration is 1.5 units.

Therefore, the correct option is A.

Note: We should be careful when we use the kinematic equation of motion. The kinematic equation of motion is used only when the acceleration is constant. If the acceleration is not constant then we have to use the definition of velocity and acceleration to form the mathematical equation. As the velocity is the rate of change of position with respect to time, so mathematically it can be written as,
\[v = \dfrac{{dx}}{{dt}}\]
As the acceleration is the rate of change of velocity with respect to time, so mathematically it can be written as,
\[a = \dfrac{{dv}}{{dt}} = \dfrac{{dv}}{{dx}} \times \dfrac{{dx}}{{dt}} = v\dfrac{{dv}}{{dx}}\]