The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. If a car traveling
50 mph can stop in 250 ft, how fast can a car travel and still stop in 90 ft?
To solve this problem, we need to use the formula for direct variation, which states that "y varies directly with x" can be expressed as y = kx, where k is the constant of variation.
In this case, the stopping distance (d) varies directly as the square of the speed (r). So we can express this relationship as:
d = k * r^2
To find the value of k, we can use the given information that a car traveling 50 mph can stop in 250 ft. Plugging in these values into the equation, we get:
250 = k * 50^2
Simplifying this equation, we have:
250 = k * 2500
Dividing both sides by 2500, we find:
k = 250/2500
k = 1/10
So now we have the constant of variation. The equation for the stopping distance can be written as:
d = (1/10) * r^2
To find the speed at which a car can still stop in 90 ft, we plug in this value for d and solve for r:
90 = (1/10) * r^2
Multiplying both sides by 10, we have:
900 = r^2
Taking the square root of both sides, we find:
r = √900
r = 30
Therefore, a car can travel at a speed of 30 mph and still stop in 90 ft.
250/2500:90/x²
225000=250x²
900=x²
x=+/-30