A scooter accelerates from rest for time $t_1$ at constant rate $a_1$ and then retards at constant rate $a_2$

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 $\frac{t_1}{t_2}$ will be:

Accepted Answer

1.  **Analyze Phase 1 (Acceleration):** The scooter starts from rest ($u=0$) and accelerates at $a_1$ for time $t_1$. Using the first equation of motion :
    $$v = u + at \Rightarrow v_{max} = 0 + a_1 t_1$$
    $$v_{max} = a_1 t_1$$
2.  **Analyze Phase 2 (Retardation):** The scooter starts with velocity $v_{max}$ and decelerates at $a_2$ (acceleration $= -a_2$) for time $t_2$ until it stops ($v_{final} = 0$).
    $$v = u + at \Rightarrow 0 = v_{max} - a_2 t_2$$
    $$v_{max} = a_2 t_2$$
3.  **Equate and Solve:** Since the maximum velocity reached is the same for both phases:
    $$a_1 t_1 = a_2 t_2$$
    $$\frac{t_1}{t_2} = \frac{a_2}{a_1}$$

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