The distance needed to stop a car (d) varies directly with the aquare of the speed of the car (s). If a car traveling 25 mph requires 0 feet to stop, how many feet will be required to stop a car that is traveling at 60 mph?
d = k v^2
when v=25, d= 0 ????
you see to have a typo .
suppose you meant to type d=50
then
50 = k(25)^2
k = 50/625 = 1/25
d = (1/25) v^2
so when v= 60
d = (1/25)(3600) = 144
Once you have determined what your typo was, replace the 50 in my solution with the correct value of d
the stopping distance D varies directly with the square of the speed S. if a car traveling 50mph has a stopping distance of 250 feet, find the stopping distance of 60mph.
What is k in this solution
300
To find out how many feet will be required to stop a car traveling at 60 mph, we need to use the direct variation relationship between distance (d) and the square of the speed (s).
The direct variation equation can be written as: d = k * s^2, where k is the constant of variation.
In this case, we are given that when the car is traveling at 25 mph, it requires 0 feet to stop. So we can substitute these values into the equation to solve for k.
0 = k * 25^2
0 = k * 625
Since anything multiplied by 0 is 0, we can conclude that k is equal to 0.
Now, we can find the distance (d) required to stop the car traveling at 60 mph:
d = k * s^2
d = 0 * 60^2
d = 0 * 3600
d = 0
Therefore, no feet will be required to stop a car traveling at 60 mph.