If v is the initial speed of a vehicle, g is the gravitational acceleration, G is the upward longitudinal slope of the road and $f_{r}$ is the coefficient of rolling friction during braking , the braking distance (measured horizontally) for the vehicle to stop is

\frac{v^{2}}{g (G + f_{r})}

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\frac{v^{2}}{2 g (G + f_{r})}

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\frac{v g}{(G + f_{r})}

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\frac{v f_{r}}{(G + g)}

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Solution

The correct option is B $\frac{v^{2}}{2 g (G + f_{r})}$
Braking distance,
$= \frac{v^{2}}{2 g (G + f_{r})}$

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Q. The coefficient of average rolling friction of a road is

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and its grade is

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. If the grade of this road is doubled, what will be the percentage change in the braking distance (for the design vehicle to come to stop) measured along the horizontal (assume all other parameters are kept unchanged)?

A vehicle is moving on a road of grade $+ 4$ % at a speed of $20 m / s$ . Consider the coefficient of rolling friction as $0.46$ and acceleration due to gravity as $10 m / s^{2}$ . On applying brakes to reach a speed of $10 m / s$ , the required braking distance (in m, round off to nearest integer) along the horizontal, is