On a two-ane road, car A is travelling with a speed of `36 km h ^(-1).` Two cars B and C approach car A in opposite directions with a speed of `54 km h ^(-1)` 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 minmum acceleration of car B is required to avoid an accident ?

Explanation

Velocity of car `A v _(A) = 36 km//hr = (36 xx 10 ^(3))/(3600) = 10 ms ^(-1)`
Velcities of car B and car C are respectively `v_(B) and v _(C) ` then
`v _(B) =v_(C) = 54 km//hr = (54 xx 10 ^(3))/(3600) = 15 ms ^(-1) `
Speed of car B w.r.t. car A,
`v_(BA)= v_(B) -v_(A) = 15 - 10 = 5 ms ^(-1)`
Speed car C w.r.t. car A.
`v _(CA) = v _(C) - v_(A)`
`= v _(C)- (-v_(A))`
` = 15 + 10 =25 ms ^(-1)`
Now, `AB=AC =1 km = 100 m.`
Suppose, time taken to cover distacne AC by car C is .t.,
`therefore d _(AC) = vt` (Because speed of car C is constnat)
`100 = 25 xx t `
`t = 40 s`
Suppose, car B covers `d _(AB) = 1000 m ` in t = 40 s with acceleratin .a.,
From equation `d= v _(0) t + 1/2 at ^(2)`
`d _(AB) = v _(BA) t + 1/2 a _(B) t ^(2)`
`1000 = 5 xx 40 + 1/2 a _(B) xx (40)^(2)`
` 1000 = 200 + 800 a _(B)`
`therefore 800 a _(B) = 800`
`a_(B) =1 ms ^(-2)`