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2.Motion in Straight Line
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An automobile travelling with a 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 ........... $m$
A$20 $
B$40$
C$60 $
D$80$
(AIEEE-2004)
Solution
d) The stopping distance, $S \propto {u^2}$
$⇒$ $\frac{{{S_2}}}{{{S_1}}} = {\left( {\frac{{{u_2}}}{{{u_1}}}} \right)^2} = {\left( {\frac{{120}}{{60}}} \right)^2} = 4$
$⇒$ ${S_2} = 4 \times {S_1} = 4 \times 20 = 80\,m$
$⇒$ $\frac{{{S_2}}}{{{S_1}}} = {\left( {\frac{{{u_2}}}{{{u_1}}}} \right)^2} = {\left( {\frac{{120}}{{60}}} \right)^2} = 4$
$⇒$ ${S_2} = 4 \times {S_1} = 4 \times 20 = 80\,m$
Standard 11
Physics
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