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 40 mph can stop in 100 ft, how fast can a car travel and still stop in 256 ft?
Where does the 5280 come from?
Just translate ...
d = k(r^2) , where k is a constant
plug in the given to find k
100/5280 = k(1600)
k = (100/1600)/(5280) = 1/84480
so your formula is
d = (1/84480)r^2
if d = 256 ...
256/5280 = (1/84480)r^2
84480(256)/5280 = r^2
256(16) = r^2
r =16(4)
= 64 mph
notice we could have skipped the 5280 ft to mile conversion, since in both cases the same units were used. e.g. mph <---> ft
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we could have used a simple ratio:
d1/d1 = (r1)^2/(r2)^2
100 ft/256 ft = (40 mph)^2/ (r2)^2
25/64 = 1600/r2^2
take √ of both sides
5/8 = 40/r2
5r2 = 320
r2 = 64
Well, isn't this a math problem. Let's see if we can put a smile on this equation, shall we?
We know that the stopping distance "d" varies directly as the square of the speed "r". So we can write this as an equation: d = kr^2, where "k" is the proportionality constant.
Now, we have two sets of values to work with. For the first case, the car is traveling at 40 mph and stops in 100 ft. Let's plug these numbers in: 100 = k * 40^2.
To find the value of "k", we need to do a little calculation. 100 = k * 1600. So, k = 100/1600 = 0.0625.
Now, we want to know how fast the car can travel and still stop in 256 ft. Let's call this speed "v". Using the same equation, we have 256 = 0.0625 * v^2.
To solve for "v", we need to get rid of 0.0625 first. Dividing both sides by 0.0625, we have v^2 = 256 / 0.0625.
Now, I'm going to create some suspense here. Prepare yourself. Take a deep breath. Are you ready? The square root of 256 / 0.0625 is... drumroll, please... 160!
So, the car can travel at a speed of 160 mph and still stop in 256 ft. That's quite a quick stop, don't you think? Just make sure there's no banana peel on the road!
To solve this problem, we need to use the concept of direct variation. The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. This can be represented as an equation: d = k * r^2, where k is the constant of variation.
We are given that a car traveling 40 mph can stop in 100 ft. Let's use this information to find the value of k. Plugging in the values we have into the equation, we get:
100 = k * (40)^2
100 = k * 1600
Now, let's solve for k:
k = 100 / 1600
k = 0.0625
We have found the value of k, which is 0.0625. Now, let's use this value to solve the remaining part of the problem.
We need to find the speed at which a car can still stop in 256 ft. Let's call this speed x. So, we have a new stopping distance (d) and need to find the corresponding speed (x). Plugging in the values into the equation, we get:
256 = 0.0625 * x^2
Let's solve for x:
256 / 0.0625 = x^2
4096 = x^2
To find x, we take the square root of both sides:
x = √4096
x = 64
Therefore, a car can travel at a speed of 64 mph and still stop in 256 ft.