Engine of a car can produce a maximum acceleration `2 m//s^(2)` and its brakes can produce a maximum retardation `3 m//s^(2)`. Minimum time in which the car can travel a distance 6 km is:

Options

A. 120 sec
B. 100 sec
C. 82 sec
D. 64 sec

Answer

B

Explanation

`s_(1) = (1)/(2) xx 2 xx t_(1)^(2)`
`6000 - s_(1) = ((2t_(1))^(2))/(3 xx 2) = (2t_(1)^(2))/(3)`
`(6000 - t_(1)^(2)) xx 3 = 2t_(1)^(2)`
`18000 = 5 t_(1)^(2)`
`rArr t_(1)^(2) = 3600`
`t_(1) = 60 sec`
Now, `s_(2) = u_(2) t_(2) - (1)/(2) a t_(2)^(2)`
`u_(2) = a_(1) t_(1)`
`2400 = 120 t_(2) - (1)/(2) xx 30 xx t_(2)^(2)`
`t_(2) = 40 sec`
Hence, total time `T = t_(1) + t_(2) = 100 sec`