You are suffering through a court case arising from a recent car accident that you were in. You were traveling West, alone in your car, which has a mass of 1000 kg, through an intersection when another driver in a small car (mass 850 kg) traveling South crashed into your passenger side at the center of the intersection. The two cars became stuck together and skidded off the road. For this problem, choose the positive x-axis to point to the East (right) and the positive y-axis to point North (up). You recall that you had just looked at the speedometer before the accident and were traveling at 20 m/s. What was your initial momentum vector?

p₀ = (- 20,000 i - 17,000 j) kg m / s

vₓ = -10.81 m / s ,   v_{y} = - 9,189 m / s  

Explanation:

This exercise can be solved with the conservation of the moment.

The system is formed by the two cars, so the forces during the crash are internal and the moment is conserved

initial moment. Just before the crash

      p₀ = M v₁ + m v₂

where the masses M = 1000 kg with a velocity of v₁ = - 20 i m / s for traveling west, the second vehicle has a mass m = 850 kg and a velocity of v₂ = - 20 i m / s, we substitute

     p₀ = - 1000 20 i - 850 20 j

     p₀ = (- 20,000 i - 17,000 j) kg m / s

final moment. Right after the crash

how the cars fit together

      = (M + m)

the final speed is shaped

      v_{f} = vₓ i + v_{y} j

  we substitute

     p_{f} = (M + m) (vₓ i + v_{y} j)

     p_{f} = (1000 + 850) (vₓ i + j)

     p₀ = p_{f}

      (- 20000 i - 17000 j) = 1850 (vₓ i + v_{y} j)

we solve for each component

    vₓ = -20000/1850

    vₓ = -10.81 m / s

    v_{y} = -17000/1850

    v_{y} = - 9,189 m / s

a.

explanation:

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14