NLM O48

Solution Q48

Question 48: Solution

1. Analysis of Forces at Start

The car starts from rest ($v=0$). At this instant, the resistive air drag force ($F_{drag} = bv$) is zero. The only horizontal force providing acceleration is the traction from the tires.

Since the wheels are “maintained always at the verge of slipping,” the traction force is the maximum static friction:

$$ F_{net} = f_s – F_{drag} $$ $$ ma = \mu mg – bv $$ At $t=0$, $v=0$: $$ ma_{initial} = \mu mg \implies a_{initial} = \mu g $$
f = μmg t = 0, v = 0 v-t Graph Calculation (0,0) (10s, 10m/s) Δt = 10 s Δv = 10

2. Calculation from Graph

The problem states the graph is linear in the first 10 seconds. We calculate the slope ($a$) using the points $(0,0)$ and $(10, 10)$:

$$ a = \frac{\Delta v}{\Delta t} = \frac{10 – 0}{10 – 0} = 1 \, \text{m/s}^2 $$

Substituting this back into our force equation (assuming $g \approx 10 \, \text{m/s}^2$):

$$ \mu g = 1 $$ $$ \mu (10) = 1 $$ $$ \mu = 0.10 $$

Correct Option: (c) 0.10