Q173 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
A particle starts from rest and moves along the x -axis with a velocity that varies as $\mathrm{v}=\sqrt{100+4 x} \mathrm{~ms}^{-1}$. The acceleration of the particle is:
- Correct answerA$2 m s^{-2}$
- B$5 m s^{-2}$
- C$25 \mathrm{~ms}^{-2}$
- D$4 m s^{-2}$
Solution
Given the velocity $v = \sqrt{100 + 4x}$.
The acceleration $a$ of a particle moving in one dimension is given by $a = v \dfrac{dv}{dx}$.
Differentiating $v = (100 + 4x)^{1/2}$ with respect to $x$:
$\dfrac{dv}{dx} = \dfrac{1}{2} (100 + 4x)^{-1/2} \times 4 = \dfrac{2}{\sqrt{100 + 4x}}$.
Substituting $v$ and $\dfrac{dv}{dx}$ into the expression for acceleration:
$a = \sqrt{100 + 4x} \times \dfrac{2}{\sqrt{100 + 4x}}$.
$a = 2 \text{ ms}^{-2}$.
Answer: $2 m s^{-2}$
The acceleration $a$ of a particle moving in one dimension is given by $a = v \dfrac{dv}{dx}$.
Differentiating $v = (100 + 4x)^{1/2}$ with respect to $x$:
$\dfrac{dv}{dx} = \dfrac{1}{2} (100 + 4x)^{-1/2} \times 4 = \dfrac{2}{\sqrt{100 + 4x}}$.
Substituting $v$ and $\dfrac{dv}{dx}$ into the expression for acceleration:
$a = \sqrt{100 + 4x} \times \dfrac{2}{\sqrt{100 + 4x}}$.
$a = 2 \text{ ms}^{-2}$.
Answer: $2 m s^{-2}$
Hence, the correct answer is option: 'A'.