As a special investigator, you are called out to the scene of an automobile cash

Question

As a special investigator, you are called out to the scene of an automobile cash. A car has skidded off a flat, dry asphalt cure, gone through the guard rail, left the edge of the embankment and become airborne (horizontally) to land on the ground below road level.

Critical Data

The radius of the curved turn is 30m

The coefficient of friction between rubber and asphalt is .8

The car's mass is 1800kg

The horizontal distance from the point of "launch" to the point of impact is 20m

The height of the embankment (the height of the vertical fall) is 10m

The guard rail is designed to absorb 4x10^5 J of energy before breaking

Questions

A. maximum safe speed of the curve?

B. what is the car's speed as if left the embankment after crashing through the guard rail?

C. what is the car's speed before striking the guard rail? (assume all energy that was lost went towards breaking the guard rail)

D. if the skid marks before the car hit the guard rail measue 2m, and the car had no ABS system, what was the speed of the car as it entered the curve?

E. when the car stopped abruptly, the occupant was restrained by the seatbelt in .5s. If the occupant was roughly 60kg, what force is the person feel?

Explanation / Answer

a) to go in the circle

friction = mv^2/r
u m g = m v^2/r

0.8*9.81 = v^2/30

v=15.34 m/s

b)

y direciton

y = y0 + v0y t + 1/2 a t^2

10 = 0.5*9.81*t^2

t=1.43 s

x direction

x = v t

v = 20/1.43= 13.99 m/s

c)

Ke initial - W = KE final
0.5*1800*v^2 -4.0E5 = 0.5*1800*13.99^2

v=25.3 m/s

d) Ei + wfriction = Ef

1/2 mv^2 - u m g d = 1/2 mv^2

0.5*1800*v^2 - 0.8*1800*9.81*2 = 0.5*1800*25.3^2
v=25.9 m/s

e)

so find v at the bottom

1/2 mv^2 + m gh = 1/2 mv^2

0.5*13.99^2 + 9.81*10 = 0.5*v^2

v=19.8 m/s

F = m v/t = 60*19.8/0.5= 2376 N

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