On a two - lane road, car A is travelling with a speed of 36 kmph. Two cars B and C approach car A in opposite direction with a speed of 54 kmph each. At a certain instant, when the distance AB is equal to AC both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?

Explanation

`v_(A)=36kmph = 36xx(5)/(18)=10ms^(-1)`
`v_(B)=v_(c)=58kmph=54xx(5)/(18)=15ms^(-1)" "vecv_(C)-vecv_(A)=15-10=+5ms^(-1)`
Time taken by car C to meet `A=(1000)/(25)=40s`
To avoid accident, the car B is accelerated such that it over takes the car A in 40 s.
`therefore x=v_(0)t+(1)/(2)at^(2), 1000=5(40)+(1)/(2)a(40)^(2), a=1ms^(-2)`