if a vehicle with a mass of 5,800lbs left skid marks that were 7ft. long, how fast were they going?
the concrete to tire coefficient of friction was .60
also, if a vehicle was 2,060lbs with skidmarks that were 20ft. long? same coefficient.
they're supposed to be going 45mph, and v2 was supposed to stop at a stop sign. did they run it?
To determine the speed at which the vehicle was traveling in both scenarios, we can use the principles of physics related to skid marks and friction. The equation we will be using is:
v^2 = u^2 + 2ad
Where:
v = final velocity (speed of the vehicle)
u = initial velocity (initial speed of the vehicle)
a = acceleration (caused by the friction)
d = distance (length of the skid marks)
Let's solve it step by step:
1. For the first scenario, where the vehicle had a mass of 5,800 lbs and left skid marks that were 7 ft. long:
First, convert the mass of the vehicle from lbs to kg. 1 lb ≈ 0.4536 kg.
Mass = 5,800 lbs × 0.4536 kg/lb ≈ 2636.08 kg
Since the coefficient of friction is given as 0.60, we can directly use it as the acceleration value.
Using the formula v^2 = u^2 + 2ad, we can rearrange it to solve for the final velocity, v.
v^2 = u^2 + 2ad
v^2 = 0 + 2 × 0.6 × 9.8 m/s^2 × 7 ft × 0.3048 m/ft
Since the initial velocity, u, is zero (assuming the vehicle started from rest before skidding), we can simplify the equation to:
v^2 = 2 × 0.6 × 9.8 m/s^2 × 7 ft × 0.3048 m/ft
Simplifying further:
v^2 = 84.4992 m^2/s^2
v ≈ √84.4992 ≈ 9.19 m/s ≈ 20.56 mph
Therefore, in the first scenario, the vehicle was approximately traveling at 20.56 mph.
2. For the second scenario, where the vehicle had a mass of 2,060 lbs and left skid marks that were 20 ft. long:
Follow the same steps as above:
Convert the mass of the vehicle from lbs to kg:
Mass = 2,060 lbs × 0.4536 kg/lb ≈ 934.64 kg
The coefficient of friction remains the same, 0.60, which we can use as the acceleration value.
Using the formula v^2 = u^2 + 2ad and assuming the initial velocity, u, is zero:
v^2 = 0 + 2 × 0.6 × 9.8 m/s^2 × 20 ft × 0.3048 m/ft
Simplifying, we get:
v^2 = 235.3216 m^2/s^2
v ≈ √235.3216 ≈ 15.33 m/s ≈ 34.33 mph
Therefore, in the second scenario, the vehicle was approximately traveling at 34.33 mph.
Regarding the last question about whether they ran the stop sign, we cannot determine that based solely on the information given. The skid marks themselves do not provide any direct evidence of running the stop sign. Additional information, such as witness accounts or camera footage, would be needed to determine if the vehicle had enough time to stop at the stop sign.