Accident investigation I know it\'s really long, but if you could nudge me towar

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Accident investigation

I know it's really long, but if you could nudge me towards the right drection, I'd be eternally grateful

D. Two-Dimensional Momentum-Accident Investigation Your consulting agency, with its extensive knowledge of physics, has been subpoenaed to provide expert testimony at an automobile accident case in civil court. The case involves a crash between a Ford Escort car and a tractor-trailer truck. 1. The police department determined that the force required to drag a 130 N (29 lb) can tire across the pavement at a constant velocity is 100 N (23 lb). Use this to find the coefficient of friction for the car tires and the road. 2. Specifications from the truck's manufacturer claim that (for technical reasons) the effective coefficient of friction for truck tires is only 70% that of car tires, what was the coefficient of kinetic friction between the tires and the road for the truck? After the collision, the truck and car skidded at the angles shown in the diagram. The car skidded a distance of 8.2 m (27 ft) before stopping while the truck skidded 11 m (37 ft) before stopping. Show that the acceleration during stopping for the vehicles is ug. Using the distance that the car skidded as it stopped and its stopping acceleration, determine the initial speed of the car. 3. Using the distance that the truck skidded as it stopped and its stopping acceleration, determine the initial speed of the car. 4.

Explanation / Answer

1)
Normal force acting = 130 N
Force required for the car to move at constant speed is same as force of friction
this force was required to overcome friction
f = 100 N

use:
f = miu* Normal force
100 = miu*130
miu = 0.77
Answer: 0.77

2)
coeffcient for truck is 70% that of car
so,
coeffcient for truck = 70/100 * coeffcient for car
=0.7 * 0.77
= 0.54
Answer: 0.54

3)
stopping acceleration = -frictional force / m = -miu*m*g/m = -miu*g
a = -miu*g
= -0.77*9.8
= -7.55 m/s^2

d = 8.2 m
Vf = 0

use:
Vf^2 = Vi^2 + 2*a*d
0 = Vi^2 + 2*(-7.55)*8.2
Vi^2 = 123.8
Vi = 11.1 m/s
Answer: initial speed of car= 11.1 m/s

4)
a = -miu*g
= -0.54*9.8
= -5.3 m/s^2

d = 11 m
Vf = 0

use:
Vf^2 = Vi^2 + 2*a*d
0 = Vi^2 + 2*(-5.3)*11
Vi^2 = 116.6
Vi = 10.8 m/s
Answer: initial speed of truck= 10.8 m/s

I am allowed to answer only 4 parts at a time

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