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 find the acceleration of a body starting from rest and moving with constant acceleration, covering a distance of \( 40 \, \text{m} \) in 4 seconds, we can use the kinematic equation:

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

Where:

• \( s \) is the distance covered, \( 40 \, \text{m} \)
• \( u \) is the initial velocity, \( 0 \, \text{m/s} \) (since the body starts from rest)
• \( a \) is the acceleration
• \( t \) is the time, \( 4 \, \text{s} \)

Substituting these values into the equation:

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

Simplifying, we get:

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

\[ 40 = 8a \]

Solving for \( a \):

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

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

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

Answer 2

To find the acceleration of a body starting from rest and moving with constant acceleration over a distance of \(40 \, \text{m}\) in \(4\) seconds, we use the kinematic equation:

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

where:

• \(s\) is the distance covered,
• \(u\) is the initial velocity,
• \(t\) is the time taken,
• \(a\) is the acceleration.

Given that the initial velocity \(u = 0\) (the body starts from rest), the equation simplifies to:

\[ s = \frac{1}{2}at^2 \]

Substituting the given values (\(s = 40 \, \text{m}\) and \(t = 4 \, \text{s}\)):

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

Calculate \(4^2 = 16\), so:

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

Simplifying further:

\[ 40 = 8a \]

Solving for \(a\):

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

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

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