A car begains from rest at time `t = 0`, and then acceleration along a straight track during the interval `0 lt t le 2 s` and thereafter with constant velocity as shown in figure in the graph. A coin is initially at rest on the floor of the car. At `t = 1 s`, the coin begains to slip and its stops slipping at `t = 3 s` . The coefficient of static friction between the floor and the coin is `(g = 10 m//s^(2))`

Options

A. `0.2`
B. `0.3`
C. `0.4`
D. `0.5`

Answer

C

Explanation

`v = 2 t^(2)`
`:. a = (dv)/(dt) = 4t` At `t = 1 s`,
`a = 4 m//s^(2)` …(i)
Limiting value of static friction ,
`f_(L) = mu _(S) mg`
`:.` Maximum value of acceleration of coin which can be provided by the friction ,
`a_(max) = (f_(L))/(m) = mu_(S) mg`
Equating Eqs. (i) and (ii), we get
`4 = mu _(S) (10)`
`mu _(S) = 0.4`