JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Kinematics

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:

(A) $\frac{a _{1} + a _{2}}{a _{1}}$
(B) $\frac{a _{2}}{a _{1}}$ ✓ Correct
(C) $\frac{a _{1} + a _{2}}{a _{2}}$
(D) $\frac{a _{1}}{a _{2}}$

Answer: (B)

Solution

The scooter accelerates from rest to velocity

v = a_{1} t_{1}

, then decelerates to rest:

0 = v - a_{2} t_{2}

. Therefore,

a_{1} t_{1} = a_{2} t_{2}

, giving

\frac{t _{1}}{t _{2}} = \frac{a _{2}}{a _{1}}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option

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