The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling 70mph can stop in 270ft, how many feet will it take the same car to stop when it is travelling 60mph?
Thank you.
To solve this problem, we need to use the concept of direct variation, which states that if two quantities are directly proportional, then their ratio is constant.
In this case, the stopping distance (d) of the car varies directly as the square of the speed (r). Mathematically, we can express this relationship as:
d ∝ r^2
We can rewrite this equation using a constant of variation (k):
d = k * r^2
To find the value of k, we can use the given information about the car's speed and stopping distance. We know that when the car is traveling at 70 mph, it can stop in 270 ft. Plugging these values into the equation, we have:
270 = k * (70)^2
Now we can solve for k:
k = 270 / (70)^2
Once we have the value of k, we can use it to find the stopping distance when the car is traveling at 60 mph. Substituting the values into the equation:
d = k * (60)^2
Now we can calculate the stopping distance:
d = k * 3600
Note that we have already calculated the value of k, so we can substitute it into this equation:
d = (270 / (70)^2) * 3600
Let's calculate this value:
d = (270 / 4900) * 3600
d ≈ 198.98 ft
Therefore, when the car is traveling at 60 mph, it will take approximately 198.98 feet to stop.