Question

When a body starts from rest and moves with uniform acceleration, then its instantaneous displacement \( s \) is related to time \( t \) as

• A. \( s \propto t^{-1} \)
• B. \( s \propto t^{1/2} \)
• C. \( s \propto t \)
• D. \( s \propto t^2 \)
• E. \( s \propto t^{-2} \)

Hint:

For motion starting from rest under constant acceleration, always remember \( s=\frac{1}{2}at^2 \), so displacement varies as the square of time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question asks for the relationship between displacement and time for an object undergoing uniformly accelerated motion, starting from rest. This is a fundamental concept in kinematics.
Step 2: Key Formula or Approach:
We use the second equation of motion for uniformly accelerated linear motion:
\[ s = ut + \frac{1}{2}at^2 \] where:
- s is the displacement.
- u is the initial velocity.
- t is the time elapsed.
- a is the uniform acceleration.
Step 3: Detailed Explanation:
The problem states that the body "starts from rest". This means the initial velocity \( u = 0 \).
It also states that the body moves with "uniform acceleration", which means the acceleration \( a \) is constant.
Let's substitute \( u = 0 \) into the equation of motion:
\[ s = (0)t + \frac{1}{2}at^2 \] \[ s = \frac{1}{2}at^2 \] In this equation, \( \frac{1}{2} \) is a constant, and \( a \) is also a constant. Therefore, the displacement \( s \) is directly proportional to the square of the time \( t \).
We can write this relationship as:
\[ s \propto t^2 \] Step 4: Final Answer:
The displacement s is related to time t as \( s \propto t^2 \). This corresponds to option (D).