non-uniform acceleration (average speed)

Question

A body moves 20 m with uniform acceleration from rest reaching speed v, then immediately 20 m with uniform deceleration coming to rest. Total time for the trip is 8 s. What was the maximum speed v reached?

RankGuru · Accepted Answer

1. **Analyze Motion Phases**: The trip consists of two symmetric segments: acceleration and deceleration.
2. **Phase 1 (Acceleration)**: Starts at $0$, ends at $v$, distance $d_1 = 20$ m.
   - Average speed in Phase 1 $= \frac{u+v}{2} = \frac{0+v}{2} = \frac{v}{2}$.
   - Time $t_1 = \frac{d_1}{v_{avg_1}} = \frac{20}{v/2} = \frac{40}{v}$.
3. **Phase 2 (Deceleration)**: Starts at $v$, ends at $0$, distance $d_2 = 20$ m.
   - Average speed $= \frac{v+0}{2} = \frac{v}{2}$.
   - Time $t_2 = \frac{20}{v/2} = \frac{40}{v}$.
4. **Total Trip Equation**: $t_1 + t_2 = 8$ s.
   - $\frac{40}{v} + \frac{40}{v} = 8$
   - $\frac{80}{v} = 8$
5. **Solve for v**: $v = \frac{80}{8} = 10$ m/s.
6. **Summary**: The maximum speed attained was **$10$ m/s**.

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non-uniform acceleration (average speed)

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