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A car begins from rest at time t=0, and then accelerates along a straight track during the interval $0 < t \leq 2 s$ and thereafter with constant velocity as shown in the graph. A coin is initially at rest on the floor of the car. At t=1s, the coin begins to slip and its stops slipping at (t=38s) coefficient of static friction between the floor and the coin is (Take, $g = 10 m / s^{2}$ )

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0.3

0.2

0.5

0.4

answer is C.

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Detailed Solution

$v = 2 t^{2}$

$a = \frac{d v}{d t} = 4 t$

$At$ t=15, $a = 4 m / s^{2}$ …………………………(i)

Limiting value of static friction.

$f_{L} = μ_{s} m g$

$∴$ Maximum value of acceleration of coin which can be provided by the friction,

$a_{\max} = \frac{f_{L}}{m} = μ_{s} g$ ………………………..(ii)

Equating Equations. (i) and (ii), we get

$4 = μ_{S} (10)$

$μ_{s} = 0.4$

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