The distance a car will skid after its brakes lock can be modelled by the equation d=kv^2, where d is the stopping distance in metres, k is a constant based on road conditions and v is the velocity of the car in metres per second at the moment the brakes lock.
1.) if a car skids 79 m when k= 0.031 how fast was the car travelling? Answer= 50m/s but I can't figure out how to get the answer.
d=kv^2
v=sqrt(d/k)=sqrt(79/.031)=sqrt(2541)=about 50 m
To find the velocity of the car, given that it skidded 79m and k = 0.031, you can use the equation d = kv^2 and solve for v.
Here's how you can do it step by step:
1. Start with the given equation: d = kv^2.
2. Plug in the values given in the question: d = 79m and k = 0.031.
3. Rewrite the equation with the provided values: 79 = 0.031v^2.
4. Divide both sides of the equation by 0.031 to isolate v^2: 79/0.031 = v^2.
5. Calculate the result: (79/0.031) = v^2.
6. Take the square root of both sides to solve for v: √(79/0.031) = v.
7. Simplify: √(2548.3871) = v.
8. Calculate the square root to find v: v ≈ 50m/s.
Therefore, the car was traveling at approximately 50 meters per second when the brakes locked.