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-uniform
• D. None

Answer 1

The correct option is A

Uniform

Explanation for the correct option(s):

A: Uniform

Given that, velocity of the particle at any point is;

$v^2=2–3x$

Differentiate above equation with respect to time we get.

$\frac{d{v}^2}{dt}=\frac{d\left(2-3x\right)}{dt}$

$2v\frac{dv}{dt}=-3\frac{dx}{dt}$

$2v\frac{dv}{dt}+3v=0$ $\left[\because velocity,v=\frac{dx}{dt}\right]$

$2\frac{dv}{dt}+3=0$

$\frac{dv}{dt}=-\frac{3}{2}$ …..$\left(i\right)$

Now,

$a=\frac{dv}{dt}$ ….$\left(ii\right)$

Where $a$ is acceleration

From equation $\left(i\right)$ and $\left(ii\right)$ we get.

$a=-\frac{3}{2}=cons\tan t$

Hence, acceleration is constant therefore it is an uniform acceleration.

Explanation for the incorrect option(s):

B: Zero

C: Non-uniform

D: None

Hence, acceleration is a constant value.

Thus, it is clear from the above proof that it is uniform.