JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Kinematics

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In a car race on straight road, car A takes a time ' t ' less than car B at the finish and passes finishing point with a speed ' v ' more than that of car B . Both the cars start from rest and travel with constant acceleration

a_{1}

and

a_{2}

respectively. Then ' v ' is equal to

(A) $\frac{2 a _{1} a _{1}}{a _{1} + a _{2}} t$
(B) $2 a_{1} a_{2} t$
(C) $a_{1} a_{2} t$ ✓ Correct
(D) $\frac{a _{1} + a _{2}}{2} t$

Answer: (C)

Solution

Both cars start from rest. Car A covers the race length

ℓ

in time

t_{A}

with

ℓ = \frac{1}{2} a_{1} t_{A}^{2}

and crosses the line at

v_{A} = a_{1} t_{A} = 2 a_{1} ℓ

; likewise car B finishes at

v_{B} = 2 a_{2} ℓ

after

t_{B}

, where

t_{A} = \frac{2 ℓ}{a _{1}}

and

t_{B} = \frac{2 ℓ}{a _{2}}

. The extra speed is

v = v_{A} - v_{B} = 2 ℓ (a_{1} - a_{2})

and the time saved is

t = t_{B} - t_{A} = 2 ℓ \frac{a _{1} - a _{2}}{a _{1} a _{2}}

. Dividing the first by the second, the common factor cancels and

\frac{v}{t} = a_{1} a_{2}

, so

v = a_{1} a_{2} t

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Kinematics

Solution

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