Distance in Constant Acceleration – Rankers Physics
Topic: Kinematics
Subtopic: Equations of Motion

Distance in Constant Acceleration

A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10\) seconds is \(S_1\) and that covered in the first \(20\)seconds is \(S_2\), then:

(2009)

\(S_2 = 3S_1\)
\(S_2 = 4S_1\)
\(S_2 = S_1\)
\(S_2 = 2S_1\)

Solution:

For constant acceleration from rest, distance \(S = \frac{1}{2}at^2\). So \(S \propto t^2\). For \(t=10 \text{ s}\), \(S_1 = \frac{1}{2}a(10)^2 = 50a\). For \(t=20 \text{ s}\), \(S_2 = \frac{1}{2}a(20)^2 = 200a\). Therefore, \(S_2 = 4S_1\).

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