Question

The speed\[v\]of a particle moving along a straight line is given by \[a + b{v^2} = {x^2}\](where \[x\]is its distance from the origin). The acceleration of the particle is
\[
  \left( 1 \right)bx \\
  \left( 2 \right)\dfrac{x}{a} \\
  \left( 3 \right)\dfrac{x}{b} \\
  \left( 4 \right)\dfrac{x}{{ab}} \\
 \]

Answer 1

Hint:In order to solve this question, we are going to first take the equation of motion given for the particle and differentiate it with respect to time. After that, the equation is simplified and then the velocity and acceleration formulae are used in order to find the acceleration of the given particle.

Formula used: The acceleration and the velocity of any particle are given by
Acceleration, \[A = \dfrac{{dv}}{{dt}}\]
Velocity,\[v = \dfrac{{dx}}{{dt}}\]
Where, \[x\]and \[t\]are displacement and time respectively.

Complete step-by-step solution:
In this question, we are given the equation of motion of the particle along a straight line. The equation of the speed of the particle moving along the straight line is given by:
\[a + b{v^2} = {x^2}\]
Differentiating both the sides of this equation with respect to time\['t'\], we get
\[0 + b\left[ {2v\dfrac{{dv}}{{dt}}} \right] = 2x\dfrac{{dx}}{{dt}}\]
Simplifying this equation, we get
\[\dfrac{{dv}}{{dt}} = \dfrac{{x\dfrac{{dx}}{{dt}}}}{{bv}}\]
Now, we know that
Acceleration, \[A = \dfrac{{dv}}{{dt}}\]
Velocity,\[v = \dfrac{{dx}}{{dt}}\]
Where, \[x\]and \[t\]are displacement and time respectively.
Using these two relations in the above equation, we get
\[A = \dfrac{{xv}}{{bv}} = \dfrac{x}{b}\]
Hence, the acceleration of the particle moving along a straight line is given by\[A = \dfrac{x}{b}\].

Note: It is important to note that when moving along a straight line, only the magnitude of the velocity i.e. the speed of the particle is changing and is taken into consideration. Also, the magnitude of the acceleration is changing in this case and not the direction, that’s why linear acceleration is taken into account.