7. A driver realizes too late that the 10 degree hill he is driving up levels of

Question

7. A driver realizes too late that the 10 degree hill he is driving up levels off abruptly as shown in the diagram. At the instant he slams on his brakes his velocity is 25 m/s and he is 15 meters from the end of the hill. The coefficient of friction between his tires and the road is 0.6. i. Find the speed of the car at the top of the hill. ii. Find the maximum height of the car above the level road. iii. Suppose that the driver had not slowed down, and had reached the level part of the road at a speed of 25 meters per second. Find the horizontal distance that the car would travel while airborne.

Explanation / Answer

initial velocity , u = 25 m/s

distance moved , S = 15 m

coefficient of kinetic friction =  ? = 0.6

acceleration = - ( g sin (10 degrees) + ? g cos (10 degress) )

   = - (9.81*sin(10 degrees) + 0.6*9.81*cos(10 degrees) )

   = - 7.5 m/s^2

v^2 - u^2 = 2aS

v^2 = u^2 + 2aS

v = sqrt(u^2 + 2aS)

v = sqrt(25^2 - 2*7.5*15) = 20 m/s

Speed at top of the hill = v = 20 m/s

ii) projectile problem

  theta = 10

Hmax = ( v sin(theta) )^2 / (2*g) = ( 20*sin(10 degrees) )^2 / (2*9.81) = 0.615 m

iii) v = 25 m/s

  Horizontal distance travelled in air = v^2 sin ( 2 theta) / g = 25^2 * sin (20 degrees) / 9.81 = 21.79 m

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Answers

i) Speed at top of the hill = 20 m/s

ii) maximum height of car = 0.615 m

iii) horizontal distance = 21.79 m

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