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

Distance Covered under Constant Acceleration

A particle has initial velocity \(2 \text{m/s}\) and has constant acceleration such that after \(2 \text{s}\) its velocity becomes \(3 \text{m/s}\). The distance travelled by particle in this duration is
\(1 \text{m}\)
\(2 \text{m}\)
\(5 \text{m}\)
\(6 \text{m}\)

Solution:

Since acceleration is constant, the distance is given by \(s = \left(\frac{u + v}{2}\right) t = \left(\frac{2 + 3}{2}
ight) \times 2 = 5 \text{m}\).

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