Question

A body starting from rest moves with a constant acceleration of \( 2 \, \text{m/s}^2 \). What is the distance covered by the body in the interval between 5 s and 6 s?

Hint:

When a body moves with constant acceleration from rest, the distance covered in a given time interval is calculated using the equation \( s = \frac{1}{2} a t^2 \).

Answer 1

Step 1: Use the equation for distance under constant acceleration.
The distance \( s \) covered by a body under constant acceleration is given by the equation: \[ s = ut + \frac{1}{2} a t^2 \] where:
- \( u \) is the initial velocity (0 m/s, since the body starts from rest),
- \( a \) is the acceleration,
- \( t \) is the time.
Since the body starts from rest, the equation simplifies to: \[ s = \frac{1}{2} a t^2 \]
Step 2: Calculate the distance covered in 6 seconds and 5 seconds.
We are given that the acceleration \( a = 2 \, \text{m/s}^2 \). The distance covered in 6 seconds is: \[ s_6 = \frac{1}{2} \times 2 \times (6)^2 = \frac{1}{2} \times 2 \times 36 = 36 \, \text{m} \] The distance covered in 5 seconds is: \[ s_5 = \frac{1}{2} \times 2 \times (5)^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{m} \]
Step 3: Find the distance covered between 5 s and 6 s.
The distance covered between 5 s and 6 s is the difference between \( s_6 \) and \( s_5 \): \[ \text{Distance between 5s and 6s} = s_6 - s_5 = 36 - 25 = 11 \, \text{m} \] Thus, the distance covered by the body between 5 s and 6 s is: \[ \boxed{11 \, \text{m}} \]