A train running at `108 km//h` is brought to a halt in 2 minutes. Calculate the retardation produced by the application of brakes. Also, calculate the distance the train travels before stopping.

Options

A. `- 0.25 m/(s^2)`, 1800 m
B. `- 0.5 m/(s^2)`, 1000 m
C. `- 1.25 m/(s^2)`, 100 m
D. NONE

Answer

A

Explanation

Here, `u = 108 km//h = (108 xx 1000m)/(60 xx 60 s) = 30 m//s`
`v = 0, t = 2 min = 2 xx 60s`
`a = ?, s = ?`
From `v = u + at`
`0 = 30 + a xx 2 xx 60`
`a = -(30)/(120) = -0.25m//s^(2)`
Negative sign indicates retardation.
From `v^(2) - u^(2) = 2as`
`0 - (30)^(2) = 2 (- 0.25)s`
`s = (-30 xx 30)/(-0.5) = 1800 m`