Question

A particle starts from rest and moves along the \(x\)-axis with a velocity that varies as \(v = \sqrt{100 + 4x}\,\text{ms}^{-1}\). The acceleration of the particle is:

• A. \(25\,\text{ms}^{-2}\)
• B. \(5\,\text{ms}^{-2}\)
• C. \(4\,\text{ms}^{-2}\)
• D. \(2\,\text{ms}^{-2}\)

Hint:

When velocity is given as function of position, use \(a = v \frac{dv}{dx}\).

Answer 1

The Correct Option is D

Step 1: Use relation between acceleration and velocity.
\[ a = v \frac{dv}{dx} \]

Step 2: Differentiate velocity.
\[ v = (100 + 4x)^{1/2} \] \[ \frac{dv}{dx} = \frac{1}{2}(100 + 4x)^{-1/2} \cdot 4 \]
\[ \frac{dv}{dx} = \frac{2}{\sqrt{100 + 4x}} \]

Step 3: Substitute into acceleration formula.
\[ a = \sqrt{100 + 4x} \cdot \frac{2}{\sqrt{100 + 4x}} \]

Step 4: Simplify.
\[ a = 2 \]

Step 5: Final conclusion.
\[ \boxed{2\,\text{ms}^{-2}} \] Hence, correct answer is option (D).