Question

What is the ratio of distances travelled by a body in the first two intervals of 5 seconds? (Given the initial velocity $ u = 1 \, \text{m/s} $ and the body moves with a constant acceleration of $ 5 \, \text{m/s}^2 $)

• A. 1:2
• B. 1:4
• C. 1:3
• D. 1:1

Hint:

The displacement increases with the square of the time due to the acceleration, so the distances travelled in successive intervals at constant acceleration will not be the same.

Answer 1

The Correct Option is B

The displacement \( s \) of a body moving with an initial velocity \( u \) and constant acceleration \( a \) is given by the equation: \[ s = ut + \frac{1}{2} a t^2 \] We need to find the ratio of the distances travelled by the body in the first and second 5-second intervals. - In the first interval (0 to 5 seconds), the displacement \( s_1 \) is: \[ s_1 = u(5) + \frac{1}{2} a (5)^2 = 1(5) + \frac{1}{2} (5) (5)^2 = 5 + \frac{1}{2} (5)(25) = 5 + 62.5 = 67.5 \, \text{m} \] - In the second interval (5 to 10 seconds), the displacement \( s_2 \) is: \[ s_2 = u(5) + \frac{1}{2} a (5)^2 = 1(5) + \frac{1}{2} (5) (5)^2 = 5 + 62.5 = 67.5 \, \text{m} \] The ratio of distances travelled in the first and second intervals is \( 1:4 \).