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Q.
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 :-
Draw $vt$ curve
$\tan \theta_{1}= a _{1}=\frac{ v _{\max }}{ t _{1}}$
$\& \tan \theta_{2}= a _{2}=\frac{ v _{\max }}{ t _{2}}$
$\div$ above
$\frac{ t _{1}}{ t _{2}}=\frac{ a _{2}}{ a _{1}}$
