Question

A particle moves in a straight line so that its velocity at any point is given $v^2=a+bx$, where $a$, $b\ne 0$ are constants. The acceleration is

• A. Zero
• B. Uniform
• C. Non-uniform
• D. Intermediate

Answer 1

The correct option is B

Uniform

Step 1: Given data.

The velocity at any point, $v^2=a+bx$

Where $a$ and $b$ are constant and $x$ is the distance covered.

Step 2: Finding the acceleration of the particle.

We know that,

$a=\frac{dv}{dt}$

Where $a$ is acceleration and $\frac{dv}{dt}$is change in velocity with respect to time.

Also,

$v=\frac{dx}{dt}$

Now,

$v^2=a+bx$ $\left[Given\right]$

Differentiate the above equation both sides with respect to time.

$\frac{d{v}^2}{dt}=\frac{d\left(a+bx\right)}{dt}$

$2v\frac{dv}{dt}=b\frac{dx}{dt}$

$2va=bv$ $\left[\because Fromequation\left(i\right)and\left(ii\right)\right]$

$a=\frac{b}{2}$

Since, it is clear that the value of acceleration $a$ is constant.

Therefore, acceleration is uniform.

Hence, option B is correct.