Q.
At he 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 $OAB$ and $CD$ respectively.
Then
A) At $t = 18$ s the motor bike and car are 180m apart
B) At $t = 18$ s the motor bike and car are 720m apart
C) The relative distance between motor bike and car reduces to zero at $t = 27$ s and both are 1080m far from origin
D) The relative distance between motor bike and car always remains same
Explanation
(c) Distance travelled by motor bike at $t = 18$s
$s_{bike} = s_{1} = \frac{1}{2}$(18)(60)=540 m
Distance travelled by car at $t = 18$s
$s_{car} = s_{2}$=(18)(60)=720 m
Therefore, separation between them at $t = 18$s is 180m. Let, separation between them decreases to zero at time $t$ beyond 18s.
Hence, $s_{bike} = 540 + 60t$ and $s_{car} = 720 + 40t$
$s_{car} - s_{bike} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ 720 + 40t = 540 + 60t$
$\Rightarrow \ \ \ \ \ \ t = 9$s beyond 18s or
Hence, $t = (18 + 9)$s=27s from start and distant travelled by both is $s_{bike}$=$s_{car} = 1080$m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.