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

Correct Answer: 100

The velocity is increasing at $$5 \text{ m s}^{-1}$$ per meter. This means:

$$\dfrac{dv}{dx} = 5 \text{ s}^{-1}$$

Acceleration is given by:

$$a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v \cdot \dfrac{dv}{dx}$$

At the point where $$v = 20 \text{ m s}^{-1}$$:

$$a = 20 \times 5 = 100 \text{ m s}^{-2}$$

Therefore, the answer is $$\boxed{100}$$.

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources