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: Recall the equation of motion.
For a body moving with uniform acceleration: \[ s = ut + \frac{1}{2}at^2 \]

Step 2: Use the condition “starts from rest”.
If the body starts from rest: \[ u = 0 \] So the equation reduces to: \[ s = \frac{1}{2}at^2 \]

Step 3: Identify the relation between \( s \) and \( t \).
From the equation: \[ s \propto t^2 \]

Step 4: Interpret physically.
With constant acceleration, displacement increases quadratically with time.

Step 5: Check dimensions.
Since \[ [a]=LT^{-2} \] we get \[ s \sim at^2 \sim L \] which is consistent.

Step 6: Verify with an example.
If \( t \) doubles: \[ s \propto t^2 \Rightarrow s \text{ becomes } 4 \text{ times} \] which matches uniformly accelerated motion.

Step 7: Final conclusion.
Hence, \[ \boxed{s \propto t^2} \] Therefore, the correct option is \[ \boxed{(4)\ s \propto t^2} \]