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

30

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Solution

The correct option is A 30
Given:
$grade = + 4$ %
$v_{1} = 20 m / s e c$
$f = 0.46$
$g = 10 m / s e c^{2}$
$v_{2} = 10 m / s e c$

We know,
$v_{2}^{2} = v_{1}^{2} + 2 a s$
$(10)^{2} = (20)^{2} + 2 s [- 10 (0.46 + \frac{4}{100})]$
$\Rightarrow s = 30 m$

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A vehicle is moving on a road of grade+4% at a speed of 20m/s. Consider the coefficient of rolling friction as 0.46 and acceleration due to gravity as 10m/s2. On applying brakes to reach a speed of 10m/s, the required braking distance (in m, round off to nearest integer) along the horizontal, is 30

The correct option is A 30Given: grade=+4% v1=20m/sec f=0.46 g=10m/sec2 v2=10m/sec We know, v22=v21+2as (10)2=(20)2+2s[−10(0.46+4100)] ⇒s=30m

The correct option is A 30
Given:
$grade = + 4$ %
$v_{1} = 20 m / s e c$
$f = 0.46$
$g = 10 m / s e c^{2}$
$v_{2} = 10 m / s e c$

We know,
$v_{2}^{2} = v_{1}^{2} + 2 a s$
$(10)^{2} = (20)^{2} + 2 s [- 10 (0.46 + \frac{4}{100})]$
$\Rightarrow s = 30 m$