JEE MAIN - Physics (2004 - No. 72)
An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 m.
If the car is going twice as fast, i.e 120 km/h, the stopping distance will be
60 m
40 m
20 m
80 m
Explanation
Assume $$a$$ be the retardation for both the vehicle then
In case of automobile,
$$u_1^2 - 2a{s_1} = 0$$
$$ \Rightarrow u_1^2 = 2a{s_1}$$
And in case for car,
$$u_2^2 = 2a{s_2}$$
$$\therefore$$ $${\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}$$
$$ \Rightarrow {\left( {{{120} \over {60}}} \right)^2} = {{{s_2}} \over {20}}$$
$$ \Rightarrow$$ s2 = 80 m
In case of automobile,
$$u_1^2 - 2a{s_1} = 0$$
$$ \Rightarrow u_1^2 = 2a{s_1}$$
And in case for car,
$$u_2^2 = 2a{s_2}$$
$$\therefore$$ $${\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}$$
$$ \Rightarrow {\left( {{{120} \over {60}}} \right)^2} = {{{s_2}} \over {20}}$$
$$ \Rightarrow$$ s2 = 80 m
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