Question
A vehicle travels at $20 m / s$ on a flat road. If the coefficient of static friction between the road and the tires is $\mu=0.4$ then find the minimum distance in which the vehicle can be stopped
✓
Answer
Given, $u=20 m / s$
Coefficient of static friction $\left(\mu_s\right)=0.4$
Let mass of vechicle $=m kg$
Reaction force of road $( R )=m g$
$\begin{array}{l}
=m \times 10 \quad \because g=10 ms^{-2} \\
=10 mN N
\end{array}$
$\begin{aligned}
\text { Friction force } & =\mu_s \times R \\
& =0.4 \times 10 m \because R=10 m \\
& =4 mN
\end{aligned}$
Deceleration due to friction force $=\frac{4 m}{m}$
$=4 m / s^2$
From third equation of motion :
$\begin{aligned}
v^2 & =u^2+2 a s \\
0 & =(20)^2+2(-4) \times s \\
8 s & =400 \\
s & =\frac{400}{8}=50 m
\end{aligned}$
Hence, minimum distance $=50 m\quad \text { Ans. }$
Coefficient of static friction $\left(\mu_s\right)=0.4$
Let mass of vechicle $=m kg$
Reaction force of road $( R )=m g$
$\begin{array}{l}
=m \times 10 \quad \because g=10 ms^{-2} \\
=10 mN N
\end{array}$
$\begin{aligned}
\text { Friction force } & =\mu_s \times R \\
& =0.4 \times 10 m \because R=10 m \\
& =4 mN
\end{aligned}$
Deceleration due to friction force $=\frac{4 m}{m}$
$=4 m / s^2$
From third equation of motion :
$\begin{aligned}
v^2 & =u^2+2 a s \\
0 & =(20)^2+2(-4) \times s \\
8 s & =400 \\
s & =\frac{400}{8}=50 m
\end{aligned}$
Hence, minimum distance $=50 m\quad \text { Ans. }$
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