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Physics-

General

Easy

Question

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by $O A B$ and $C D$ respectively Then

At $t = 18$ s the motor bike and car are 180m apart
At $t = 18$ s the motor bike and car are 720m apart
The relative distance between motor bike and car reduces to zero at $t = 27$ s and both are 1080m far from origin
The relative distance between motor bike and car always remains same

The correct answer is: The relative distance between motor bike and car reduces to zero at $t equals 27$ s and both are 1080m far from origin

Distance travelled by motor bike at $t equals 18$ s
$s subscript b i k e end subscript equals s subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction$ (18)(60)=540 m
Distance travelled by car at $t equals 18$ s
$s subscript c a r end subscript equals s subscript 2 end subscript$ =(18)(60)=720 m
Therefore, separation between them at $t equals 18$ s is 180m. Let, separation between them decreases to zero at time $t$ beyond 18s.
Hence, $s subscript b i k e end subscript equals 540 plus 60 t$ and $s subscript c a r end subscript equals 720 plus 40 t$
$s subscript c a r end subscript minus s subscript b i k e end subscript equals 0$
$rightwards double arrow blank 720 plus 40 t equals 540 plus 60 t$
$rightwards double arrow blank t equals 9$ s beyond 18s or
Hence, $t equals open parentheses 18 plus 9 close parentheses$ s=27s from start and distant travelled by both is $s subscript b i k e end subscript$ = $s subscript c a r end subscript equals 1080$ m

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