Two cars `C_(1)` and `C_(2)` moving in the same direction on a straight single lane road with velocities `v_(1)=12 m s^(-1)` and `v_(2) =10 m s^(-1)`, respectively . When the separation between the two was `d=200 m`, `C_(2)` started accelerating to avoid collision. What is the minimum acceleration of car `C_(2)` so that they do not collide?
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Explanation

Acceleration is just avoided if the relative velocity becomes zero just at the moment the two cars meet each other, i.e.
`veca_(2) =vec a_(1)-vec a_(c_(1)- vec a_(c)_(2) =0-a(-a)`
`vec u_(12) =vec u_(1)-vec u_(2) =12-10 =2 ms^(-2)`
The collsion is just avoided if the relative velocity becomes zero just at the moment the two cars meer each other, i.e.,
`v_(12)=0` when `s_(12)=200`.
Now `v_(12)=0, vec u_(12)=2, =vec a_(12)=-a`, and `s_(12) =200`.
`v_(12)^(2)-u_(12)^(2)=2a_(12)s_(12) becuase 0-2^(2)=-2xxaxx200`
`rarr a-(1)/(100)ms^(-2) =0. 1 ms^(-2) =1 cms^(-2)`
Therefore, minimum acceleration needed by car `c_(2)=1 cm s^(-2)`.