Question

The distance travelled by a vehicle at speed V on an upgrade after the brake is applied is equal to the distance travelled by the same vehicle at 0.75 V on down grade of same amount. The grade will be :
(Take f=0.30)

• A. 4.8%
• B. 6.8%
• C. 8.4%
• D. 9.6%

Answer 1

The correct option is C 8.4%

Braking distance = $\frac{V^2}{2g(f\pm n)}$
From equation,
$\frac{V^2}{2g(f+n)}=\frac{(0.75V{)}^2}{2g(f-n)}$
$\frac{f+n}{f-n}=\frac{16}{9}$
$\frac{f+n+(f-n)}{f+n-(f-n)}=\frac{16+9}{16-9}[Bycomponendo-dividendo]$
$\frac{f}{n}=\frac{25}{7}$
$n=\frac{7}{25}\times 0.3=0.084=8.4$%