Question

The driver of vehicle travelling at 65 kmph up a grade takes 10 m less distance to stop after he applies the brake than a driver travelling at same initial speed down the grade. If the coefficient of friction is 0.40, then the gradient is _____ percent.

• A. 4.74
• B. 7.11
• C. 2.37
• D. 3.38

Answer 1

The correct option is A 4.74

As, vehicle travelling towards upgrades takes 10 m less than travelling towards downgrade with same grade, so
$\frac{V^2}{254(f+n)}=\frac{V^2}{254(f-n)}-10$
$\Rightarrow \frac{{65}^2}{254(0.4+n)}=\frac{{65}^2}{254(0.4-n)}-10$
$\Rightarrow 16.63(\frac{1}{0.4-n}-\frac{1}{0.4+n})=10$
$\Rightarrow 3.33n={0.4}^2-n^2$
$\Rightarrow n^2+3.33n-{0.4}^2=0$
$\Rightarrow n=0.0474$
So, gradient of pavement is $4.74$%