SOLUTION: a. Suppose two animals are in a race. Choose the two animals and calculate the speed of each animal in yards per second. b. Decide how much of a head start (in yards) the faster

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Question 611365: a. Suppose two animals are in a race. Choose the two animals and calculate the speed of each animal in yards per second.
b. Decide how much of a head start (in yards) the faster animal offers the slower animal. For each animal, write an equation relating distance from the starting line to time.
c. Graph the two equations on the same coordinate plane
d. How long will it take the faster animal to overtake the slower animal? How many yards from the starting line are the animals when the faster animal overtakes the slower animal?
e. Reduce the head start by half and repeat parts c and d.
So I did a and I picked an Ostrich 45 mph and a lion 50 mph. I converted them and got 22 yards per second for the ostrich and 24.44 yards per second for the lion.
The rest of the questions have me confused. Please tell me how to do these and provide an explanation I'm really stuck on this and it's due in two days.
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

For part b, pick a number, any number. Pick something that you can divide by 2 easily since in step e you are going to cut the head start in half.

Let represent time in seconds. Then an equation for any animal is

Where is the speed of the animal in yards per second and is the head start. So for your example, the ostrich is modeled by:

where is whatever you picked for a value in part b. And for the Lion:

Notice that the Lion is the fast animal so in the Lion's equation is zero.

Graph the equations. The solution set consists of the point where the two lines intersect.

The first part of part d is the value of the independent variable at the point where the two graphs intersect. The second part of part d is the value of the function at that value of the independent variable.

You can check your graphical answer by setting the two functions equal to each other and then solving for and then substituting that value of back in to either function to determine the value of the function at that point.

Part e is just more of the same.

John

My calculator said it, I believe it, that settles it