JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 20)
A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2 . The car will catch up with the bus after a time of :
$$\sqrt {110} \,s$$
$$\sqrt {120} \,s$$
$$10\,\,\sqrt 2 \,s$$
15 s
Explanation
Acceleration of Car, aC = 4 m/s2
Acceleration of bus, aB = 2 m/s2
Initial distance between them, S = 200 m
Acceleration of Car with respect to bus,
aCB = aC $$-$$ aB = 4 $$-$$ 2 = 2 m/s2
As initially both are at rest so, uCB = 0
$$\therefore\,\,\,$$ S = UCB $$ \times $$ t + $${1 \over 2}$$ aCB $$ \times $$ t2
$$ \Rightarrow $$$$\,\,\,$$ 200 = 0 + $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ t2
$$ \Rightarrow $$$$\,\,\,$$ t2 = 200
$$ \Rightarrow $$$$\,\,\,$$ t = 10$$\sqrt 2 $$ sec.
Acceleration of bus, aB = 2 m/s2
Initial distance between them, S = 200 m
Acceleration of Car with respect to bus,
aCB = aC $$-$$ aB = 4 $$-$$ 2 = 2 m/s2
As initially both are at rest so, uCB = 0
$$\therefore\,\,\,$$ S = UCB $$ \times $$ t + $${1 \over 2}$$ aCB $$ \times $$ t2
$$ \Rightarrow $$$$\,\,\,$$ 200 = 0 + $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ t2
$$ \Rightarrow $$$$\,\,\,$$ t2 = 200
$$ \Rightarrow $$$$\,\,\,$$ t = 10$$\sqrt 2 $$ sec.
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