A scooter accelerates from rest for time t1 at a constant rate -1 and then retards at constant rate -2 for time t2 and comes to rest- The correct value of t1t2 will be -

During acceleration, v=0+α_1t_1=α_1t_1 During retardation, 0=v α_2t_2 ⇒ v=α_2t_2 ⇒α_1t_1=α_2t_2 ⇒t_1t_2=α_2α_1 Hence, (B) is the correct answer. ...

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Byju's Answer

Standard XI

Physics

Applications of equations of motion

A scooter acc...

Question

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

\frac{α_{1} + α_{2}}{α_{2}}

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\frac{α_{2}}{α_{1}}

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\frac{α_{1} + α_{2}}{α_{1}}

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\frac{α_{1}}{α_{2}}

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Solution

The correct option is B $\frac{α_{2}}{α_{1}}$
During acceleration,

$v = 0 + α_{1} t_{1} = α_{1} t_{1}$

During retardation,

$0 = v - α_{2} t_{2}$

$\Rightarrow v = α_{2} t_{2}$

$\Rightarrow α_{1} t_{1} = α_{2} t_{2}$

$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{α_{2}}{α_{1}}$

Hence, $(B)$ is the correct answer.

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A scooter accelerates from rest for time t1 at a constant rate α1 and then retards at constant rate α2 for time t2 and comes to rest. The correct value of t1t2 will be -

The correct option is B α2α1During acceleration, v=0+α1t1=α1t1 During retardation, 0=v−α2t2 ⇒v=α2t2 ⇒α1t1=α2t2 ⇒t1t2=α2α1 Hence, (B) is the correct answer.

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