The stopping distance d of a car after tehe brakes are applied varies directly as the square of the speed r. If a car traveling 50 mph can stop in 140 ft. How many feet will it take the same car to stop when it is traveling 70 mph?
I did everything and I got 6860/25, how do I convert that to lowest terms, they are not looking for a mixed fraction, which is the only thing I know how to do. Thank you
r=140 (70/50)^2= 140*49/25=274 ft
thanks, but they said to use form,140=k*50^2, then I got 140=k(2500) which ends up being k=14/250. The equation variation is d=14/250*r^2 which give us d=14/250*70^2 or 6860/25. I just didn't know how to convert to lowest terms. Can you help me with this? Other examples I got wrong are 9800/36 I was suppose to convert to 2450/9, I don't know how they got this, do you?
To find the answer, we first need to understand the relationship between the stopping distance and speed. The problem states that the stopping distance, d, varies directly with the square of the speed, r. This can be represented as:
d = k * r^2
Where k is the constant of variation.
Now, we are given that when the car is traveling at 50 mph, it can stop in 140 ft. So we can substitute these values into the equation:
140 = k * 50^2
Now, let's solve for k:
k = 140 / 2500
k = 0.056
Now that we have the value of k, we can use it to find the stopping distance when the car is traveling at 70 mph. Substituting the new speed into the equation:
d = 0.056 * 70^2
To calculate this, let's start by squaring 70:
d = 0.056 * 4900
d = 274.4
So, the car will take approximately 274.4 feet to stop when it is traveling at 70 mph.
Now, to convert the fraction 6860/25 to lowest terms:
Step 1: Find the greatest common divisor (GCD) of the numerator and denominator, which is 5 in this case.
Step 2: Divide both the numerator and the denominator by the GCD:
6860 ÷ 5 = 1372
25 ÷ 5 = 5
So, the fraction 6860/25 in lowest terms is 1372/5.