Question

The displacement of a particle at the time \( t \) is given by \[ s = \sqrt{1 + t}, \quad \text{then its acceleration } a \text{ is proportional to} \]

• A. square of the velocity
• B. \( \sqrt{s} \)
• C. \( \sqrt[3]{s} \)
• D. cube of the velocity

Hint:

To find the relationship between velocity and acceleration, first compute the velocity and then differentiate to find the acceleration.

Answer 1

The Correct Option is D

Step 1: Find the velocity.
The velocity \( v \) is the first derivative of displacement with respect to time: \[ v = \frac{ds}{dt} = \frac{d}{dt} \left( \sqrt{1 + t} \right) = \frac{1}{2\sqrt{1 + t}} \cdot 1 \] Thus, \[ v = \frac{1}{2\sqrt{1 + t}}. \]
Step 2: Find the acceleration.
The acceleration \( a \) is the derivative of velocity with respect to time: \[ a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{1}{2\sqrt{1 + t}} \right) = -\frac{1}{4(1 + t)^{3/2}}. \]
Step 3: Relating acceleration to velocity.
We can see that the acceleration is proportional to the cube of the velocity, as shown by the relationship between \( v \) and \( a \). Therefore, the acceleration is proportional to the cube of the velocity. Thus, the correct answer is option (D).