You testify as a \"expert witness\" in a case involving an accident in which car
ID: 1590696 • Letter: Y
Question
You testify as a "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is theta = 12 degrees...that the cars were separated by distance d = 26.5 m when the driver of the car A put the car into a slide (it lacked any automatic anti-brake-lock-system), and that the speed of car A at the onset of braking was v(initial) = 16.0 m/s.
(1) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
(2) What was the speed if the coefficient of kinetic friction was 0,10 (road surface covered with wet leaves)?
Explanation / Answer
The forces on car A are the component of the weight down the slope, m g sin(12) = 2.04 * m, and the force of friction. f. If the friction coefficient is 0.6, the value of f is (0.6)mg cos(12) = 5.868 m,
directed up the slope,
so the total force is 2.04 m - 5.868 m = - 3.828 m (minus means directed up the slope).
Newton's law ; -3.828 m = m a, so a = -3.828 m/s^2.
The equation for distance travelled with constant acceleration is
d = 26.5 = vo t1- 0.5(3.828) t1^2, where vo = 16 m/s.
Solving the quadratic equation for t1 gives t1 = 2.275 s
(ignoring 6.083 as it gives wrong velocity).
The velocity at the time of collision is vo - 3.828(2.275) = 7.29 m/s.
Part -2
If the coefficient of friction is 0.10, the value of f is
(0.1)mg cos(12) = 0.95 * m, directed up the slope,
so the total force is 2.04 m - 0.95 m = 1.09 m DIRECTED DOWN THE SLOPE.
In this case the acceleration (down the slope) is 1.09 m/s^2.
The equation for distance travelled with constant accelleration is
d = 26.5 = vo t2 + 0.5(1.09) t2^2, where vo = 16 m/s.
Solving the quadratic equation for t2 gives t2 = 1.572 s (ignoring the other root as the value is negative).
The velocity at the time of collision is vo + 1.09(1.572) = 17.713 m/s
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