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Two cars 1 and 2 move with velocity v 1 and v 2 respectively on a straight road in the same direction. When the cars are separated by distance d, the driver of car 1 applies brakes and the car moves with uniform retardation a 1 . Simultaneously, car 2 starts accelerating with a 2 . If v 1 >v 2 , find the minimum initial seperation between the cars to avoid collision between them. While doing this, i reached at the equation: d = (v 1 - v 2 ) 2 - (v' 1 - v' 2 ) 2 / 2(a 1 + a 2 ) After this it is given in solution that d is max when v' 1 = v' 2 ... can you explain this relation..? Or suggest me some better alternative method..

Two cars 1 and 2 move with velocity v1 and v2 respectively on a straight road in the same direction. When the cars are separated by distance d, the driver of car 1 applies brakes and the car moves with uniform retardation a1. Simultaneously, car 2 starts accelerating with a2. If v1>v2, find the minimum initial seperation between the cars to avoid collision between them.

While doing this, i reached at the equation:

d = (v1 - v2)2 - (v'1 - v'2)2 / 2(a1 + a2)

After this it is given in solution that d is max when v'1 = v'2... can you explain this relation..?

Or suggest me some better alternative method..

Grade:12

1 Answers

Manas Shukla
102 Points
7 years ago
Since you have already arrived at the equation I will assume it as correct.
Now the equation u mentioned.
d is a fraction and will be maximum when numerator is maximum.
Now if we have a term Z = X-Y , it will be maximum when Y = 0.
Thus in your equation when (v'1 - v'2)2 =0 
and thus v'1 = v'2

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