A scooter accelerates from rest for time t_{1 | vedclass

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(B) Let $v_{\max}$ be the maximum velocity attained by the scooter.During the acceleration phase,the final velocity is $v_{\max} = 0 + a_{1}t_{1}$,so

Vedclass · Accepted Answer

$\frac{a_{2}}{a_{1}}$

Vedclass · Answer

(B) Let $v_{\max}$ be the maximum velocity attained by the scooter.During the acceleration phase,the final velocity is $v_{\max} = 0 + a_{1}t_{1}$,so $t_{1} = \frac{v_{\max}}{a_{1}}$.During the retardation phase,the final velocity is $0 = v_{\max} - a_{2}t_{2}$,so $t_{2} = \frac{v_{\max}}{a_{2}}$.Taking the ratio of the two times:$\frac{t_{1}}{t_{2}} = \frac{v_{\max} / a_{1}}{v_{\max} / a_{2}} = \frac{a_{2}}{a_{1}}$.Alternatively,using the velocity-time graph,the slope of the acceleration part is $a_{1} = 	an 	heta_{1} = \frac{v_{\max}}{t_{1}}$ and the magnitude of the slope of the retardation part is $a_{2} = 	an 	heta_{2} = \frac{v_{\max}}{t_{2}}$.Thus,$\frac{t_{1}}{t_{2}} = \frac{a_{2}}{a_{1}}$.

$A$ scooter accelerates from rest for time $t_{1}$ at a constant rate $a_{1}$ and then retards at a constant rate $a_{2}$ for time $t_{2}$ and comes to rest. The correct value of $\frac{t_{1}}{t_{2}}$ will be ..... .

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