Question

A body starts from rest and moves with an acceleration of \(2 \, \text{m/s}^2\) and comes to rest with a retardation of \(2 \, \text{m/s}^2\). Find the distance travelled?

• A. \(20 \, \text{m}\)
• B. \(25 \, \text{m}\)
• C. \(30 \, \text{m}\)
• D. \(40 \, \text{m}\)

Hint:

When an object accelerates and then decelerates symmetrically, the total distance travelled is twice the distance travelled during either phase.

Answer 1

The Correct Option is C

First, let's calculate the distance travelled during acceleration. The equation of motion is: \[ v^2 = u^2 + 2a s \] Where: - \(v\) is the final velocity, - \(u\) is the initial velocity, - \(a\) is the acceleration, - \(s\) is the distance. For the first part of the motion, the body starts from rest, so \(u = 0\), and the acceleration \(a = 2 \, \text{m/s}^2\). Let \(s_1\) be the distance travelled during acceleration. Using the equation of motion: \[ v^2 = 0 + 2 \times 2 \times s_1 \] \[ v^2 = 4 s_1 \] For the second part, the body comes to rest, so the final velocity \(v = 0\) and the retardation \(a = -2 \, \text{m/s}^2\). Let \(s_2\) be the distance travelled during retardation. Using the equation of motion: \[ 0 = v^2 - 2 \times 2 \times s_2 \] \[ v^2 = 4 s_2 \] Since the velocity is the same for both parts of the motion, we have: \[ 4 s_1 = 4 s_2 \quad \Rightarrow \quad s_1 = s_2 \] Thus, the total distance travelled is: \[ s_{\text{total}} = s_1 + s_2 = 2 s_1 \] Substituting the values: \[ s_1 = 15 \, \text{m} \quad \Rightarrow \quad s_{\text{total}} = 30 \, \text{m} \] Thus, the total distance travelled is 30 meters.