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Standard XII

Mathematics

nth Term of A.P

A car is goin...

Question

A car is going at a speed of $6 m / s$ , when it encounters a $15 m$ slope of angle $30^{o}$ . The friction coefficient between the road and tyre is $0.5$ . The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is ( $g = 10 m / s^{2}$ )

4 m / s

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3 m / s

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7.49 m / s

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8.45 m / s

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Solution

The correct option is C $7.49 m / s$
$f = μ m g c o s 30$
$a_{d o w n} = g s i n 30 - μ g c o s 30 = g (\frac{1}{2} - \frac{0.5 \sqrt{3}}{2}) = 5 - 2.5 \sqrt{3} = 0.66 m / s^{2} v^{2} = u^{2} + 2 a_{d o w n} s 36 + 2 \times 0.66 \times 15 v = 7.49 m / s$

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A car is going at a speed of 6m/s, when it encounters a 15m slope of angle 30o. The friction coefficient between the road and tyre is 0.5. The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is (g=10m/s2)

The correct option is C 7.49m/sf=μmgcos30adown=gsin30−μgcos30=g(12−0.5√32)=5−2.5√3=0.66 m/s2v2=u2+2adowns36+2×0.66×15v=7.49m/s

A car is going at a speed of $6 m / s$ , when it encounters a $15 m$ slope of angle $30^{o}$ . The friction coefficient between the road and tyre is $0.5$ . The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is ( $g = 10 m / s^{2}$ )

The correct option is C $7.49 m / s$
$f = μ m g c o s 30$
$a_{d o w n} = g s i n 30 - μ g c o s 30 = g (\frac{1}{2} - \frac{0.5 \sqrt{3}}{2}) = 5 - 2.5 \sqrt{3} = 0.66 m / s^{2} v^{2} = u^{2} + 2 a_{d o w n} s 36 + 2 \times 0.66 \times 15 v = 7.49 m / s$