Distance Covered under Constant Acceleration – Rankers Physics

Equations of Motion: Practice Problem & Solution

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 Explained:

To solve this problem, we apply the core principles of Equations of Motion. Understanding the underlying formula is key to arriving at the correct answer below:

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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Similar Practice Questions

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1. While sitting on a tree branch \(20\text{m}\) above the ground, you drop a walnut. When the walnut has fallen \(5\text{m}\) you throws a second walnut straight down. What initial speed must you give the second walnut if they are both to reach the ground at the same time? (g=\(10\text{m/s}^2\))
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