disregard the previous I did not include the first part
This question is typical on some driver's license exams: A car moving at 50 km skids 15 m with locked brakes.How far will the car skid with locked brakes at 120 km/h?
work done by friction = F d
F is friction force, assume the same in both cases
F d = (1/2) m v^2
F(15) = (1/2) m (50)^2
F d = (1/2) m (120)^2
d=(1/2) m (120)^2 (15)/ [(1/2)m (50)^2 ]
d = 15 (120)^2/(50)^2
To solve this problem, we need to use the concept of proportional reasoning. The distance the car skids is directly proportional to the square of its speed. We can express this relationship using the formula:
(d1 / d2) = (v1^2 / v2^2)
Where:
- d1 is the distance the car skids at the first speed
- d2 is the distance the car will skid at the second speed
- v1 is the first speed
- v2 is the second speed
Let's plug in the values given in the question:
d1 = 15 m (skid distance at 50 km/h)
v1 = 50 km/h
v2 = 120 km/h (the speed we need to find the skid distance for)
Now we can rearrange the formula to solve for d2:
d2 = d1 * (v2^2 / v1^2)
Substituting the values, we get:
d2 = 15 m * (120^2 km/h) / (50^2 km/h)
Now, let's calculate the solution step by step:
1. First, square the values in the numerator and denominator:
d2 = 15 m * (14400 km^2/h^2) / (2500 km^2/h^2)
2. Next, simplify the expression inside the parentheses:
d2 = 15 m * (0.056 km^2/h^2)
3. Multiply the two values:
d2 = 0.84 m
Therefore, the car will skid approximately 0.84 meters with locked brakes at a speed of 120 km/h.