Your friend has just been in a traffic accident and hopes that you can show the

Question

Your friend has just been in a traffic accident and hopes that you can show the accident was the other drivers fault. Your friends car was traveling North when it entered the intersection. When it reached the center of the intersection, the car was struck by the other drivers car which was traveling East. The two cars remained joined together after the collision and skidded to a stop. The speed limit on both roads is 50mph. From the skid marks still visible on the street, you determine that after the collision the cars skidded 56 feet at an angle of 30 degrees north of east before stopping. The police report gives the make and year of each car. The weight of your friends car is 2600 lbs and that of the other car is 2200 lbs, including the drivers weight in each case. The coefficient of kinetic friction for a rubber tire skidding on dry pavement is 0.80. You decide to see if the other driver was speeding and if your friend was under the speed limit.

How fast was your friend driving?

How fast was the other driver driving?

Neat writing is appreciated! Thank you,

Explanation / Answer

Given

speed limit of roads at intersection    = 50 mph = 50*0.44704

                                                           = 22.352 m/s

weight of her   w1 = 2600 lbs

                             = 11565.37 N

mass of her car   m1 = 1179.33 kg

weight of other car   w2 = 2200 lbs = 9786.0875 N

mass pf other car   m2 = 997.90 kg

after the collision the cars skidded 56 feet at an angle of 30 degress north of east before stopping

According to law of conservation of momentum

              m1 u1 = ( m1 + m2 )v cos    ...........(1)

              1179.33 *u1 = ( 1179.33 + 997.90) v cos 30  

                m 2 u2 = ( m1 + m2 ) v sin 30 ...........(2)

after the collision the cars skidded 56 feet  north of east before stopping  

      From kinematic equations

                  v '2 - v 2 = 2 as     

The coefficient of kinetic friction for a rubber tire skidding on dry pavement is 0.8

                  v '2 - v 2 = 2 as      

      after collision initial speed of   cars is

                 - v2 = 2 ( -g ) s

                           = 2(0.8*9.8) (17.06 m )

                    v       = 16.355 m /s

initial speed of her car is calculated from equation (1)

              1179.33 *u1 = ( 1179.33 + 997.90) (16.355m/s)cos 30

                         u1 = ( 1179.33 + 997.90) (16.355m/s)cos 30 / 1179.33

                               = 26.1487 m /s

initial speed of other car

997.90 *u1 = ( 1179.33 + 997.90)(16.355 m/s ) sin 30   

                         u1 =   ( 1179.33 + 997.90)(16.355 m/s ) sin 30 / 997.90

                               = 17.8417m /s

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