A particle is moving in a straight line such that its velocity is increasing at 5 ms ⁻¹ per meter. The acceleration of the particle is _____________ ms ⁻² at a point where its velocity is 20 ms ⁻¹. Numerical answer type. Work it out, then open the answer.

Answer: 100. Velocity increases at 5 m/s per metre, so dv/dx = 5 s⁻¹ a = vdv/dx = 20 × 5 So a = 100 m/s².

A particle is moving in a straight line such that its velocity is… | JEE Main 2022 (25 Jul, Shift 2)

Q22JEE Main 2022 · 25 JUL · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

A particle is moving in a straight line such that its velocity is increasing at 5 ms ${}^{-1}$ per meter. The acceleration of the particle is _____________ ms ${}^{-2}$ at a point where its velocity is 20 ms ${}^{-1}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 100

Velocity increases at $5$ m/s per metre, so $\dfrac{dv}{dx} = 5\text{ s}^{-1}$
$a = v\dfrac{dv}{dx} = 20 \times 5$
So $a = 100\text{ m/s}^2$ .

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