Question

The driver of vehicles traveling at 60 kmph up a gradient requires 9 m less to stop after he applies brakes, as compared to a driver traveling at the same speed, down the same gradient. Given f = 0.40. What is the percent gradient?

Answer 1

$Given,V=60kmph,f=0.40$.
Breaking distance up the gradient,
$S_1=\frac{(0.278\times V)}{2g(f+n)}$
$=\frac{(0.278\times {60}^2)}{2\times 9.81(0.40+n)}$

Braking distance down the gradient
$S_2=\frac{(0.278\times V{)}^2}{2g(f-n)}=\frac{(0.278\times 60{)}^2}{2\times 9.81(0.4-n)}$

Given,
$S_1=S_2-9$
$\Rightarrow S_1-S_2=-9m$
$\Rightarrow \frac{(0.278\times {60}^2}{2\times 9.81}\times [\frac{1}{0.4+n}-\frac{1}{0.4-n}]=-9$

$\text{Solving for}n$
$n=0.05ie.,$ $5\%$