Time for Constant Acceleration – Rankers Physics
Topic: Kinematics
Subtopic: Equations of Motion

Time for Constant Acceleration

A particle moves in a straight line with a constant acceleration. It changes its velocity from \(10 \text{ m s}^{-1}\) to \(20 \text{ m s}^{-1}\) while passing through a distance \(135 \text{ m}\) in (t) second. The value of (t) is

(2008)

12
9
10
1.8

Solution:

Given \(u = 10 \text{ m/s}\), \(v = 20 \text{ m/s}\), (\S = 135 \text{ m}\). Using \(v^2 = u^2 + 2aS\), \((20)^2 = (10)^2 + 2a(135) \Rightarrow 400 = 100 + 270a \Rightarrow a = \frac{300}{270} = \frac{10}{9} \text{ m/s}^2\). Now use \(v = u + at\), \(20 = 10 + \frac{10}{9}t \Rightarrow 10 = \frac{10}{9}t \Rightarrow t = 9 \text{ s}\).

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