Question

A particle moves in a straight line in such a way that its velocity at any point is given by \[{v^2} = 2 - 3x\], where \[x\] is measured from a fixed point. The acceleration is
A. uniform
B. zero
C. non-zero
D. indeterminate

Answer 1

Hint: Use the formula for the acceleration of the particle in terms of velocity and velocity of the particle in terms of displacement of the particle. Take the differentiation of the velocity of the particle with respect to time to determine the acceleration of the particle.

Formula Used: The acceleration \[a\] of an object is given by
\[a = \dfrac{{dv}}{{dt}}\] …… (1)

Here, \[dv\] is the change in velocity of the object during the time interval \[dt\].

The velocity \[v\] of an object is given by
\[v = \dfrac{{dx}}{{dt}}\] …… (2)

Here, \[dx\] is the change in displacement of the object during the time interval \[dt\].

Complete step by step answer:
The particle is moving in a straight line with velocity \[v\].
\[{v^2} = 2 - 3x\]

Take square root on both sides of the above equation.
\[v = \sqrt {2 - 3x} \]

Determine the acceleration of the particle.

Substitute \[\sqrt {2 - 3x} \] for \[v\] in equation (1).
\[a = \dfrac{{d\left( {\sqrt {2 - 3x} } \right)}}{{dt}}\]

Divide and multiply by \[dx\] on the right hand side of the above equation.
\[a = \dfrac{{d\left( {\sqrt {2 - 3x} } \right)}}{{dx}}\dfrac{{dx}}{{dt}}\]

Substitute \[v\] for \[\dfrac{{dx}}{{dt}}\] in the above equation.
\[a = \dfrac{{d\left( {\sqrt {2 - 3x} } \right)}}{{dx}}v\]
\[ \Rightarrow a = v\left[ {\dfrac{1}{2}{{\left( {2 - 3x} \right)}^{ - \dfrac{1}{2}}}\left( {\dfrac{{d\left( {2 - 3x} \right)}}{{dx}}} \right)} \right]\]
\[ \Rightarrow a = v\left[ {\dfrac{1}{2}\dfrac{1}{{\sqrt {2 - 3x} }}\left( {0 - 3} \right)} \right]\]

Substitute \[\sqrt {2 - 3x} \] for \[v\]in the above equation.
\[a = - \sqrt {2 - 3x} \left[ {\dfrac{3}{2}\dfrac{1}{{\sqrt {2 - 3x} }}} \right]\]
\[ \Rightarrow a = - \dfrac{3}{2}\,{\text{unit}}\]

Therefore, the acceleration of the particle is \[ - \dfrac{3}{2}\,{\text{unit}}\].

This acceleration of the particle is constant and remains the same at all times.

Therefore, the acceleration of the particle is uniform.

Hence, the correct option is A.

Note:The negative sign of the acceleration indicates that the particle has a negative acceleration i.e. retardation which shows that the velocity of the particle is decreasing. But the particle has a constant and hence uniform negative acceleration. Also the unit of the acceleration is not mentioned in the answer as the unit of velocity or displacement is not given. The particle has the acceleration according to the units of the velocity and displacement.