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# A migrating elephant herd started moving at a rate of 6 miles per hour. One elephant stood still and was left behind. Then this stray elepha

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A migrating elephant herd started moving at a rate of 6 miles per hour. One elephant stood still and was left behind. Then this stray elephant sensed danger and began running at a rate of 10 miles per hour to reach the herd. The stray caught up in 5 minutes.

18. How long (in hours) did the stray run to catch up? How far did it run?

19. Find the distance that the herd traveled while the stray ran to catch up. Then write an expression for the total distance the herd traveled. Let x represent the distance (in miles) that the herd traveled while the stray elephant stood still.

20. Use the distances you found in Exercises 18 and 19. Write and solve an equation to find how far the herd traveled while the stray stood still.

math
Aug 19, 2014

#1
+13

Every minute, the herd travels 1/10 of a mile. And every minute, the stray travels 1/6 of a mile.

18. It took the stray 1/12 of an hour to catch up. Note that, if the stray ran for 5 miniutes, the distance it would cover would have been (R * T)  =  (1/6)mi/min * 5 min = 5/6 mile.

19. Note that, in the five minutes it took the stray to catch up, the herd traveled R * T = x = (1/10) mi/min * 5 min = x = 1/2 mile

20. Note that, if the stray ran 5/6 of a mile, the herd  must have covered (5/6 - 1/2) = 1/3 of a mile while the stray stood still. To see why this is so, note that the herd travels (1/3) of a mile while the stray rests, and then travels another (1/2) mile in the five minutes the stray takes to catch up. So..... (1/3)mi + (1/2)mi = (5/6) mi. And that's how far the stray ran in the five minutes. In essence, the stray makes up (1/3) of a mile in 5 minutes. And this is so because his rate of (1/6)mi/min is being "retarded" by the (1/10)mi/min that the herd is traveling. So, his "efffective" rate  is just (1/6 - 1/10) = 1/15mi/min. And multiplying this by 5 min = 1/3 mi.

So our Equation in this part is:  (1/6 -1/10)t = d    where t is the time it takes the stray elephant to catch the herd once it starts running and d is the distance the herd traveled while the stray stood still.

So

(1/6 - 1/10)mi/min *(5)min = 1/3 mi

(1/15)mi/min * (5 min) = 1/3 mi.   Aug 19, 2014

#1
+13

Every minute, the herd travels 1/10 of a mile. And every minute, the stray travels 1/6 of a mile.

18. It took the stray 1/12 of an hour to catch up. Note that, if the stray ran for 5 miniutes, the distance it would cover would have been (R * T)  =  (1/6)mi/min * 5 min = 5/6 mile.

19. Note that, in the five minutes it took the stray to catch up, the herd traveled R * T = x = (1/10) mi/min * 5 min = x = 1/2 mile

20. Note that, if the stray ran 5/6 of a mile, the herd  must have covered (5/6 - 1/2) = 1/3 of a mile while the stray stood still. To see why this is so, note that the herd travels (1/3) of a mile while the stray rests, and then travels another (1/2) mile in the five minutes the stray takes to catch up. So..... (1/3)mi + (1/2)mi = (5/6) mi. And that's how far the stray ran in the five minutes. In essence, the stray makes up (1/3) of a mile in 5 minutes. And this is so because his rate of (1/6)mi/min is being "retarded" by the (1/10)mi/min that the herd is traveling. So, his "efffective" rate  is just (1/6 - 1/10) = 1/15mi/min. And multiplying this by 5 min = 1/3 mi.

So our Equation in this part is:  (1/6 -1/10)t = d    where t is the time it takes the stray elephant to catch the herd once it starts running and d is the distance the herd traveled while the stray stood still.

So

(1/6 - 1/10)mi/min *(5)min = 1/3 mi

(1/15)mi/min * (5 min) = 1/3 mi.   CPhill Aug 19, 2014
#2
+8

Chris pointed out that this was a really good question and I just want to try and work through it myself.

I find this type of question quite challenging to get my head around. This one is a little easier because the questions are ordered and lead you in the right direction. 18) How long (in hours) did the stray run to catch up? How far did it run?

The straggler ran for five minutes at 10miles per hour.

5 minutes is   $$\frac{1}{12}$$   of an hour

$$\\\frac{10miles}{hour}\times \frac{1}{12} hour\\\\ =\frac{10miles}{hour}\times \frac{1hour}{12} \\\\ \mbox{The hours cancel out}\\\\ =10miles\times \frac{1}{12} \\\\ =\frac{10}{12}miles \\\\ =\frac{5}{6}miles \\\\$$

19) and 20)

The herd was travelling at 6m/h.  The straggler ran for   $$\frac{1}{12}$$  of an hour

In this time the herd travelled       $$6\times \frac{1}{12}=\frac{1}{2}mile=0.5miles$$

Then write an expression for the total distance the herd traveled.

Let x represent the distance (in miles) that the herd traveled while the stray elephant stood still.

$$\\x+\frac{1}{2}=\frac{5}{6}\\\\ x=\frac{5}{6}-\frac{1}{2}\\\\ x=\frac{2}{6}\\\\ x=\frac{1}{3}mile\\\\$$

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Aug 21, 2014