Question

A body starts from rest and moves with constant acceleration. If it covers a distance of $ 40 \, \text{m} $ in 4 seconds, then its acceleration is:

• A. \( 5 \, \text{m/s}^2 \)
• B. \( 2.5 \, \text{m/s}^2 \)
• C. \( 10 \, \text{m/s}^2 \)
• D. \( 4 \, \text{m/s}^2 \)

Hint:

Key Fact: For uniformly accelerated motion from rest: \( s = \frac{1}{2} a t^2 \)

Answer 1

The Correct Option is A

To determine the acceleration of an object that begins at rest and moves with uniform acceleration, traversing a distance of \( 40 \, \text{m} \) over 4 seconds, the following kinematic equation is applied:

\[ s = ut + \frac{1}{2} a t^2 \]

Key parameters are:

• \( s \) = distance = \( 40 \, \text{m} \)
• \( u \) = initial velocity = \( 0 \, \text{m/s} \) (starting from rest)
• \( a \) = acceleration
• \( t \) = time = \( 4 \, \text{s} \)

Substituting these known values into the equation yields:

\[ 40 = 0 \times 4 + \frac{1}{2} a (4^2) \]

The simplified equation is:

\[ 40 = \frac{1}{2} a \times 16 \]

\[ 40 = 8a \]

Solving for \( a \):

\[ a = \frac{40}{8} \]

\[ a = 5 \, \text{m/s}^2 \]

Therefore, the body's acceleration is \( 5 \, \text{m/s}^2 \).