Question:
A scooter accelerates from rest for time $t_{1}$ at constant rate $\mathrm{a}_{1}$ 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 :-
Correct Option: , 2
Solution:
Draw vt curve
$\tan \theta_{1}=a_{1}=\frac{v_{\max }}{t_{1}}$
$\& \tan \theta_{2}=\mathrm{a}_{2}=\frac{\mathrm{v}_{\max }}{\mathrm{t}_{2}}$
$\div$ above
$\frac{t_{1}}{t_{2}}=\frac{a_{2}}{a_{1}}$
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