Amy was driving on a rural, two lane paved road and turned north onto a narrow gravel road with which she was unfamiliar. After about .25 mile, the road made about a 45 degrees turn to the right followed almost immediately by a 45 degrees turn to the left and then by a narrow wooden bridge spanning a small creek. Tree limbs and brush were encroaching on the road, diminishing the visibility of the second turn. Amy negotiated the first turn, but decided she wasn't going to make the second turn and started braking.
Due to the low coefficient of friction on the gravel road (mu = 0.45 ), the brakes locked on the car and she lost directional control. Skidmarks indicated a skid distance "s" = 32.7 ft before she went airborne off the bank of the creek. The front of her car hit the far bank of the creek. Measurements utilizing the impact scar on the far bank indicated that the center of mass of the car traveled a horizontal distance L = 44.6 ft and a vertical distance h = 8.3 ft downward during the time "?" that it was airborne. The approach path to the bank prior to going airborne was approximately straight and level.
One item of interest in analyzing this accident is to determine the speed Amy was driving prior to beginning the skid. How do we determine the speed when we have incomplete skidmarks? We can set the work done by friction during skidding equal to the change in kinetic energy during skidding:
Wmus = 0.5WV12/g - 0.5WV22/g
Where V1 is the speed prior to braking, V2 is the speed leaving the bank, W is the loaded weight, and g = 32.2 ft/sec/sec is the gravitational constant. Thus, we could calculate V1 if we knew V2, since all the other variables are known, but how do we determine V2? There is more than one way. But we must determine the best and fastest method.
The best way is to use methods for determining ballistic trajectories in physics. Even though the car is not a point mass, it may be considered a point mass concentrated at the center of mass for the purpose of this calculation. Of course, measurements should be made with consideration to the dimensions of the car. We should also realize that after the front wheels go airborne, the force of gravity and the force on the rear wheels produce a small downward pitching angular velocity of the car, so that it will not be horizontal when it impacts the far bank. Since aerodynamic drag is small, there is no significant horizontal force during the short time the car is airborne. Thus, the horizontal component of velocity is essentially constant (V2) while airborne. Since the distance of horizontal travel is equal to the velocity times the time, we have:
L = V2 ? or ? = L/V2
Also, while airborne, the only significant vertical force is that due to gravity. With a constant downward acceleration of "g," the vertical distance traveled is:
"h" = 0.5g?2 = 0.5g(L/V2)2,
using the expression for T obtained above. Thus, we can solve for V2.
V2 = L(g/2h)1/2 = 42.4 mph
Using this result in the first equation above, we solve for V1:
V1 =47.3 mph.
Fortunately, Amy had on her seat belt and survived with bruises and sprains. She had to wear dark glasses for several days.