Kinematics

JEE Main level, single correct, medium

64 of 308

Question 64 of 308

An automobile, traveling at 40kmh−140km{h}^{-1}, can be stopped at a distance of 40m40m by applying brakes. If the same automobile is traveling at 80kmh−180km{h}^{-1}, the minimum stopping distance in metres is (Assume no skidding):

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Answer: (D) 160m160m

S=u22aS=\dfrac{{u}^{2}}{2a}
S1S2=u12u22⇒S2=(u2u1)2S1=(2)2(40)=160m\dfrac{{S}_{1}}{{S}_{2}}=\dfrac{{u}_{1}^{2}}{{u}_{2}^{2}}\Rightarrow {S}_{2}={\left(\dfrac{{u}_{2}}{{u}_{1}}\right)}^{2}{S}_{1}={\left(2\right)}^{2}\left(40\right)=160m

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