The length of skid marks a car makes when braking varies directly with the square of the speed of the car. If a car traveling 25 mph leaves skid marks 30 ft long, how long would the skid marks be if the car is traveling 40 mph??
x = k v^2
30 = k (625)
k = 30/625
x = (30/625)(1600)
= 76.8 ft
Hey Amy, you should be able to do these now.
yea im getting it now that one I did by myself and just needed to know if I got it right thanks for all the help :)
To solve this problem, we can use the concept of direct variation. Direct variation states that two variables are directly proportional to each other if one variable is a constant multiple of the other.
Let's denote the length of skid marks by 'L' and the speed of the car by 'S'.
According to the problem, the length of skid marks 'L' varies directly with the square of the speed 'S'. Mathematically, we can express this as:
L = k * S^2
where 'k' is the constant of variation.
We are given that when the car is traveling at 25 mph, the skid marks are 30 ft long. Using this information, we can substitute the values into the equation to solve for 'k'.
30 = k * 25^2
30 = k * 625
k = 30 / 625
k = 0.048
Now that we have the value of 'k', we can use it in the equation to find the length of skid marks when the car is traveling at 40 mph.
L = 0.048 * 40^2
L = 0.048 * 1600
L = 76.8 ft
Therefore, when the car is traveling at 40 mph, the skid marks would be approximately 76.8 feet long.