Question:
A scooter accelerates from rest for time \(t_1\) at constant rate \(a_1\) and then retards at constant rate \(a_2\) for time \(t_2\) and comes to rest. The correct value of \(\frac{t_1}{t_2}\) wil be :
A scooter accelerates from rest for time \(t_1\) at constant rate \(a_1\) and then retards at constant rate \(a_2\) for time \(t_2\) and comes to rest. The correct value of \(\frac{t_1}{t_2}\) wil be :
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The speed gained during acceleration equals the speed lost during retardation, so \(a_1 t_1 = a_2 t_2\).
Updated On: Nov 4, 2026
- \(\frac{a_1 + a_2}{a_2}\)
- \(\frac{a_1 + a_2}{a_1}\)
- \(\frac{a_2}{a_1}\)
- \(\frac{a_1}{a_2}\)
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The Correct Option is C
Solution and Explanation
Maximum speed reached after time \(t_1\): \(v = 0 + a_1 t_1\).
During retardation the scooter goes from \(v\) to 0: \(0 = v - a_2 t_2\), so \(v = a_2 t_2\).
Hence \(a_1 t_1 = a_2 t_2\), giving \(\frac{t_1}{t_2} = \frac{a_2}{a_1}\).
Answer: (3) \(\frac{a_2}{a_1}\)
During retardation the scooter goes from \(v\) to 0: \(0 = v - a_2 t_2\), so \(v = a_2 t_2\).
Hence \(a_1 t_1 = a_2 t_2\), giving \(\frac{t_1}{t_2} = \frac{a_2}{a_1}\).
Answer: (3) \(\frac{a_2}{a_1}\)
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