Question:
A particle is moving along a straight line such that its velocity is increasing at $5\text{ ms}^{-1}$ per meter. When its velocity becomes $20\text{ ms}^{-1}$ its acceleration is:
A particle is moving along a straight line such that its velocity is increasing at $5\text{ ms}^{-1}$ per meter. When its velocity becomes $20\text{ ms}^{-1}$ its acceleration is:
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Remember that acceleration has two primary mathematical expressions:
$a = \frac{dv}{dt}$ (with respect to time) and
$a = v\frac{dv}{dx}$ (with respect to position).
Use the spatial derivative form when the rate is given "per meter".
$a = \frac{dv}{dt}$ (with respect to time) and
$a = v\frac{dv}{dx}$ (with respect to position).
Use the spatial derivative form when the rate is given "per meter".
Updated On: Jul 22, 2026
- $50\text{ ms}^{-2}$
- $75\text{ ms}^{-2}$
- $100\text{ ms}^{-2}$
- $10\text{ ms}^{-2}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We need to find the instantaneous acceleration of a particle given its velocity and the rate of change of its velocity with respect to position.
Step 2: Key Formula and Approach:
The standard formula for acceleration $a$ as a function of position $x$ is:
\[ a = v \frac{dv}{dx} \] where $v$ is the velocity and $\frac{dv}{dx}$ is the velocity gradient (change in velocity per unit distance).
Step 3: Detailed Explanation:
• Identify given values:
The velocity gradient $\frac{dv}{dx}$ is $5\text{ ms}^{-1}\text{ per meter} = 5\text{ s}^{-1}$.
The instantaneous velocity $v$ is $20\text{ ms}^{-1}$.
• Calculate acceleration:
Using the position-derived acceleration formula:
\[ a = v \frac{dv}{dx} \] Substitute the values into the equation:
\[ a = 20\text{ ms}^{-1} \times 5\text{ s}^{-1} \] \[ a = 100\text{ ms}^{-2} \]
Step 4: Final Answer:
The acceleration of the particle is $100\text{ ms}^{-2}$, which corresponds to Option (C).
We need to find the instantaneous acceleration of a particle given its velocity and the rate of change of its velocity with respect to position.
Step 2: Key Formula and Approach:
The standard formula for acceleration $a$ as a function of position $x$ is:
\[ a = v \frac{dv}{dx} \] where $v$ is the velocity and $\frac{dv}{dx}$ is the velocity gradient (change in velocity per unit distance).
Step 3: Detailed Explanation:
• Identify given values:
The velocity gradient $\frac{dv}{dx}$ is $5\text{ ms}^{-1}\text{ per meter} = 5\text{ s}^{-1}$.
The instantaneous velocity $v$ is $20\text{ ms}^{-1}$.
• Calculate acceleration:
Using the position-derived acceleration formula:
\[ a = v \frac{dv}{dx} \] Substitute the values into the equation:
\[ a = 20\text{ ms}^{-1} \times 5\text{ s}^{-1} \] \[ a = 100\text{ ms}^{-2} \]
Step 4: Final Answer:
The acceleration of the particle is $100\text{ ms}^{-2}$, which corresponds to Option (C).
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