Question

While driving at a speed of 30 kmph (with available friction 0.4) down the grade, the driver requires a braking distance twice that required for stopping the vehicle when he travels up the same grade. The grade is

Answer 1

Braking distance for travelling upgrade,
$S_1=\frac{V^2}{254(f+0.01n)}$

Braking distance for travelling downgrade
$S_2=\frac{V^2}{254(f-0.01n)}$

$Given,S_1=2S_1$
$\Rightarrow \frac{1}{(f-0.01n)}=\frac{2}{(f+0.01n)}$
$\Rightarrow f+0.01n=2f-0.02n$
$\therefore n=\frac{1}{0.03}=\frac{0.4}{0.03}=13.3\%$