Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

\frac{a_1 + a_2}{a_2}

\frac{a_2}{a_1}

\frac{a_1}{a_2}

\frac{a_1 + a_2}{a_1}

Step-by-Step Solution

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 : $v = u + at \Rightarrow v_{max} = 0 + a_1 t_1$ $v_{max} = a_1 t_1$
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 ( $v_{final} = 0$ ). $v = u + at \Rightarrow 0 = v_{max} - a_2 t_2$ $v_{max} = a_2 t_2$
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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