-vehicle 1: 2,060lbs

-vehicle 2: 5,800lbs
-both roads are 45mph
-The cars were impacted at right angles
-roadway was dry, 69 degrees
-friction between tire & road: .60=mu
-skid marks (PRIOR to impact): v1=20ft, v2=7ft.

*you can find the entire problem w/ more info online if you google: "john henry vehicle 5,800" it'll be the first one.

we need to figure out what was the
-velocity of both cars before impact.
-how fast does a car need to go to leave a 7ft. skid mark from a dead stop? (this ties into trying to solve who was speeding, or if both were, and if one ran the stop sign.)

*please show all work and formulas used. thankyou!

To solve the problem, we will use the principle of conservation of momentum and the formula for skid marks. Let's begin by finding the velocity of both cars before the impact.

1. Velocity of Vehicle 1 (v1) before impact:
Since Vehicle 1 leaves a skid mark of 20ft, we can use the formula for skid marks:

v1^2 / (30 * mu * g) = d1

Where v1 is the initial velocity of Vehicle 1, mu (μ) is the coefficient of friction between the tire and road (given as 0.60), g is the acceleration due to gravity (approximately 32.17 ft/s^2), and d1 is the skid mark length (20ft).

Plugging in the values:

v1^2 / (30 * 0.60 * 32.17) = 20

Now, solve for v1 by rearranging the equation:

v1^2 = 20 * 30 * 0.60 * 32.17

v1^2 ≈ 7747.712

Taking the square root of both sides:

v1 ≈ √7747.712

v1 ≈ 88.01 ft/s

Therefore, the velocity of Vehicle 1 before the impact was approximately 88.01 ft/s.

2. Velocity of Vehicle 2 (v2) before impact:
Since Vehicle 2 leaves a skid mark of 7ft, we can use the same formula:

v2^2 / (30 * mu * g) = d2

Where v2 is the initial velocity of Vehicle 2, mu (μ) is the coefficient of friction between the tire and road (given as 0.60), g is the acceleration due to gravity (approximately 32.17 ft/s^2), and d2 is the skid mark length (7ft).

Plugging in the values:

v2^2 / (30 * 0.60 * 32.17) = 7

Now, solve for v2 by rearranging the equation:

v2^2 = 7 * 30 * 0.60 * 32.17

v2^2 ≈ 1620.036

Taking the square root of both sides:

v2 ≈ √1620.036

v2 ≈ 40.25 ft/s

Therefore, the velocity of Vehicle 2 before the impact was approximately 40.25 ft/s.

Now, let's determine how fast a car would need to go to leave a 7ft skid mark from a dead stop.

Using the same formula as before:

v^2 / (30 * mu * g) = d

Where v is the required velocity, mu (μ) is the coefficient of friction between the tire and road, g is the acceleration due to gravity (approximately 32.17ft/s^2), and d is the skid mark length (7ft).

Plugging in the values:

v^2 / (30 * 0.60 * 32.17) = 7

Now, solve for v by rearranging the equation:

v^2 = 7 * 30 * 0.60 * 32.17

v^2 ≈ 1620.036

Taking the square root of both sides:

v ≈ √1620.036

v ≈ 40.25 ft/s

Therefore, a car would need to go at a velocity of approximately 40.25 ft/s to leave a 7ft skid mark from a dead stop.

Please note that this answer assumes ideal conditions and doesn't take factors such as vehicle weight and other variables into account. It's always best to consult with accident reconstruction experts or professionals for a detailed analysis and accurate conclusions.