Physics

Physics

Problem

A particle starts from rest at the origin and moves eastward with a constant acceleration of 3 m s−23\,\text{m}\,\text{s}^{-2} for 4 s4\,\text{s}. Immediately afterwards, its acceleration reverses direction but keeps the same magnitude for the next 4 s4\,\text{s}. East is maintained as the positive direction throughout. What is the particle’s velocity at t=8 st=8\,\text{s}? a) +24 m s−1+24\,\text{m}\,\text{s}^{-1} b) 0 m s−10\,\text{m}\,\text{s}^{-1} c) +12 m s−1+12\,\text{m}\,\text{s}^{-1} d) −12 m s−1-12\,\text{m}\,\text{s}^{-1}
Intuitive explanation

The key is to track how acceleration changes the velocity in two separate phases. During the first 4 seconds, constant acceleration builds up eastward velocity from rest. Then the acceleration flips direction—now opposing the motion—so it removes that velocity just as quickly. Since both phases have equal magnitude and duration, the gains and losses are symmetric, leaving you to think about what net effect remains.

Intuitive explanation

The key is to track how acceleration changes the velocity in two separate phases. During the first 4 seconds, constant acceleration builds up eastward velocity from rest. Then the acceleration flips direction—now opposing the motion—so it removes that velocity just as quickly. Since both phases have equal magnitude and duration, the gains and losses are symmetric, leaving you to think about what net effect remains.

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