Question

The driver of a car requires 20 m less to stop after he applies the brakes while travelling up a grade than the driver travelling at the same initial speed down the same grade. Consider the coefficient of friction between tyres and pavement as 0.35 and initial speed to be 80 kmph. The grade (in percentage) will be

• A. 5.12
• B. 2.38
• C. 2.56
• D. 4.77

Answer 1

The correct option is D 4.77

$\text{Braking distance}=\frac{V^2}{2g(f\pm n\%)}$

$=\frac{V^2}{254(f\pm n\%)}$

n = grade of pavement
+ve for upgrade
-ve for downgrade

f = 0.35, V = Speed of car = 80 kmph
As vehicle travelling upgrade requires 20 m less than travelling downgrade.

$So,\frac{V^2}{254(f+n)}=\frac{V^2}{254(f-n)}-20$

$\Rightarrow \frac{{\left(80\right)}^2}{254(0.35+n)}=\frac{{\left(80\right)}^2}{254(0.35-n)}-20$

$\Rightarrow \frac{{\left(80\right)}^2}{254}\left[\frac{0.35-n-0.35-n}{{\left(0.35\right)}^2-n^2}\right]=-20$

$\Rightarrow \frac{-2n}{{\left(0.35\right)}^2-n^2}=-0.79375$

$\Rightarrow 0.79375n^2+2n-0.09723=0$

$\therefore n=0.0477$

$\therefore \text{Grade}\left(n\%\right)=4.77\%$