SOLUTION: Suppose that the temperature outside is dropping at a constant rate. At noon, the temperature is 70oF and it drops to 58oF at 5:00 pm. How much did the temperature drop each hour
Question 76226: Suppose that the temperature outside is dropping at a constant rate. At noon, the temperature is 70oF and it drops to 58oF at 5:00 pm. How much did the temperature drop each hour? Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! At noon the temperature was 70 degrees F. Five hours later (at 5:00 p.m.) the temperature
was 58 degrees F. So in five hours the temperature dropped 70 - 58 or 12 degrees.
.
Since the rate of drop was constant, all you have to do is divide the total amount of drop
(12 degrees F) by the number of hours that it took to drop that amount (5 hours) and you
get the rate of drop (call it R). In equation form this is:
.
R = 12/5 = 2+(2/5) = 2.4 degrees F per hour.
.
Hope this discussion helps you to see how to calculate the constant rate of drop that
you were asked to find. If you want to check the temperature at each hour you could
start at 70 degrees and subtract 2.4 degrees each hour. If you do you will find that
at 1:00 p.m. the temperature should be 67.6 degrees. By 2:00 p.m. it has lost another
2.4 degrees so it is 67.6 - 2.4 = 65.2 degrees. By 3:00 p.m it has lost another 2.4 degrees
so the temperature is 65.2 - 2.4 = 62.8 degrees. Then by 4:00 p.m. the temperature
is down another 2.4 degrees so it is 62.8 - 2.4 = 60.4 degrees. And an hour later
(5:00 p.m.) it has dropped another 2.4 degrees and is 60.4 - 2.4 = 58 degrees.
.
That's exactly the temperature that the problem says it should be at 5:00 p.m. so our
answer appears to be right.
(