JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 4)
A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for time t2 and comes to rest. The correct value of $${{{t_1}} \over {{t_2}}}$$ wil be :
$${{{a_1} + {a_2}} \over {{a_2}}}$$
$${{{a_1} + {a_2}} \over {{a_1}}}$$
$${{{a_2}} \over {{a_1}}}$$
$${{{a_1}} \over {{a_2}}}$$
Explanation
_26th_February_Evening_Shift_en_4_1.png)
From given information :
For 1st interval
$${a_1} = {{{v_0}} \over {{t_1}}}$$
v0 = a1 t1 ....... (1)
For 2nd interval
$${a_2} = {{{v_0}} \over {{t_2}}}$$
v0 = a2 t2 ..... (2)
from (1) & (2)
a1 t1 = a2 t2
$${{{t_1}} \over {{t_2}}} = {{{a_2}} \over {{a_1}}}$$
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