At the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both…

At 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 OAB and CD respectively. Then                     
  1. At t=18 s the motor bike and car are 280m apart
  2. At t=18 s the motor bike and car are 720m apart
  3. The relative distance between motor bike and car reduces to zero at t=27 s and both are 1080m far from origin
  4. The relative distance between motor bike and car always remains same

Solution

Distance travelled by motor bike at t=18s

sbike=s1=ut +12at2u=0⇒ s1=12at=12×18×60=540m

Distance travelled by car moving with uniform velocity at t=18s

scar=s2=ut=18×60=720m (a=0)

Therefore, separation between them at t=18s is 180m.

Let, separation between them decreases to zero at time t beyond 18s.

Hence, Displacement of bike in 27 sec is  sbike=540+60t=540 +60×9=1080m

 and displacement of car in 27 sec is scar=40t=40×27=1080m

scar=sbike .

Asked in: JEE Mains - Motion In One Dimension - Test 4

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