Two cars `1` and `2` move with velocities `v_(1)` and `v_(2)`, respectively, on a straight road in same direction When the crs are separated by a destance `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) lt v_(2)`, find the minimum initial separation between the cars to avoid collision between then.

Explanation

Method 1: Let assume that the cars udergo displacements `s_(1)` and `s_(2)` aftre a time `t` from the instant of beaking.
From , `s_(1)-s_(2)=d` …(i)
where `s_(1)=v_(1)t-(1)/(2) a_(1)t^(2)` (ii)
.
and `s_(2)t+(1)/(2)a_(1)t^(2)` ...(iii)
Using (i), and (ii), we have
`(v_(1)t-(1)/(2) a_(1)t^(2))-(v_(2)t+(1)/(2)+(1)/(2)a_(1)t_(2))=d`
or `(a_(1)+a_(2))t^(2)-2)(v_(1)-v_(2))t+2d=0`
Which is quadratic equation in `t`, solving for `t`
This gives `t=(v_(1)-v_(2)+-sqrt((v_(1)-v_(2))^(2)-2(a_(1)(a1+a_(2)) (d)/(a_(1)+as_(2))`
If cars do not collede, `b^(2)-4ac lt 0`
`((v_(1)-v_(2))^(2))/(2(a_(1)+a_(2))`
Hence, `d_(min)=((v_(1)-v_(2)))^(2)/(2(a_(1)+a_(2))`
Method 2: Relative velocity apparoach
Using `v_(ret)^(2)=u_(ret)^(2)+2a_(rel).s_(rel)` We n obtain the same result.
Substitute: `v_(rel)=v'_(12)=v'_(12)-v'_(2), u_(rel)=v_(1)-v_(2)`,
`a_(rel)=a_(12)=(-a_(1))-a_(1)=-(a_(1)+a_(2))` and `s_(rel)=s_(12) =-d`,
to obtaion `(v'_(1)-v'_(2)=(v_(1)-v_(2))^(2)-2(a(a_(1)+a_(2))d`
Since `v'_(1)ltv'_(2)` to avoid collision, we have
`d=((v_(1)-v_(2)^(2)-(v'_(21)-v'_(2))^(2)/(2(a_(1)+a_(2))`
Maximum distance (d) for collistion is possible when `v'_(1)=v`_(2),
the, `d_(max)=((v_(1)-v_(2)))^(2)`.