Distance under Uniform Acceleration – Rankers Physics
Topic: Kinematics
Subtopic: Equations of Motion

Distance under Uniform Acceleration

If a car at rest accelerates uniformly to a speed of \(144 \text{ km/h}\) in \(20 \text{ sec}\), it covers a distance of:

(1997)

\(1440 \text{ cm}\)
\(2980 \text{ cm}\)
\(20 \text{ m}\)
\(400 \text{ m}\)

Solution:

Given (u=0), \(v = 144 \text{ km/h} = 144 \times \frac{5}{18} = 40 \text{ m/s}\), \(t = 20 \text{ s}\). Using \(S = \frac{u+v}{2}t), we get \(S = \frac{0+40}{2} \times 20 = 20 \times 20 = 400 \text{ m}\).

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