Question

A body starts from rest and is moving with a constant acceleration \( a \). The relation between instantaneous displacement and time is

• A. \( s = ut + \frac{1}{2} at^2 \)
• B. \( s = \frac{1}{2} at^2 \)
• C. \( s = a t^2 \)
• D. \( s = v t \)

Hint:

For a body starting from rest, the displacement in uniformly accelerated motion is given by \( s = \frac{1}{2} at^2 \), where \( a \) is the constant acceleration.

Answer 1

The Correct Option is A

For a body starting from rest and moving with a constant acceleration \( a \), the equation for displacement \( s \) is given by the equation of motion for uniformly accelerated motion: \[ s = ut + \frac{1}{2} at^2 \] where: - \( s \) is the displacement, - \( u \) is the initial velocity (which is zero in this case because the body starts from rest), - \( a \) is the acceleration, - \( t \) is the time. Since the body starts from rest, \( u = 0 \), so the equation simplifies to: \[ s = \frac{1}{2} at^2 \] Thus, the relation between displacement and time is \( s = \frac{1}{2} at^2 \), which corresponds to option (B). The correct answer is \( s = \frac{1}{2} at^2 \).