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`        two cars are moving in same direction.1st car is accelerating constantly from rest at 2 m/s^2 and 2nd car is moving with velocity of 8m/s.(left to right)the distance between 1st and 2nd car is 50m.At what time will the cars meet and at what distance from their  initial positions?`
one year ago

Shailendra Kumar Sharma
188 Points
```							Car first started with 2m/s2 whereas 2nd car is moving with 8m/sNow 50m is the distace between the carso the relative velocity b/w both of them is 8m/s and relative accelaration is 2m/s2s=ut+ ½ at2 50= 8t+(1/2)2t2t2+8t-50=0=> t2+10t-2t-50=0=>t=2,-10so time will be 2 sec
```
one year ago
Shivam
12 Points
```							If u don`t want to do with relative motion u can also do this question by this methodMark two car one is behind the other name them A and B Let A is first car and B is second car.A is moving with constant acceleration of 2m/s^2. And B is moving with velocity 8 m/s.Let a point C where they meet  and meet after t second s.Displacement of A = 50m + displacement of B .Displacement of A = ut +1/2 at^2 As A is moving with constant acceleration then u = 0 for A. Now,Displacement of A = 0 +1/2 ×2×t^2=t^2.Displacement of B = ut +1/2at^2 acceleration for B will be zero.Displacement of B = 8 ×t +0 =8tAs I tell above disp. Of A = 50 + disp. Of B t^2 = 50+ 8tt^2 -8t - 50 =0By solving get ur answer.For second part Put the value of t in displacement of AandB u get ur second answer
```
one year ago
Shivam
12 Points
```							I have put space between lines but it is showing different and in answer 1 factor r wrong in equation of t
```
one year ago
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