Expert: Bobby Soltani - 8/11/2004

Question
I'm very sorry, but I copied the question wrong. Billy's average rate
is 50 mph, not 200 mph. I should have been more careful!  

So, it must be:
x1 = (t-5)*56
x2 = 215 - (t-6)*50
then
56t - 280 = 215 - 50t - 300
56t + 50t = 280 + 215 - 300
    106t = 195
       t = 1.8..
So the time they are passing each other is 1:08? Can not be...there must be a mistake in my calculation?



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Followup To
Question -
But the answer is 7:30pm. How so?

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Followup To
Question -
Hi Bobby and thank you for your reply.
So, t is the time the second plane overtakes the first one and not the time (or hours) that took for the second plane to overtake the first one.
I thought t had to be always the amount of time that took to cover a certain distance in the formula D=Speed*Time and I did not know it could also represent the time of departure.

Then, I guess it's possible to apply that to the following question:

**At 5 p.m., Jill leaves New York City and travels toward Boston at an average rate of 56 m.p.h. At 6 p.m., Billy leaves Boston and travels toward New York, on the same road as Jill, at an average rate of 200 m.p.h. If they were initially 215 miles apart, at what time do Billy and Jill pass each other?**

Answer:
1. Set up equations that represent the distance both Jill and Billy will be travelling with respect to time.
215=56*(t-5) (Jill)
215=50*(t-6) (Billy)

2. Distance is represented as..
215= 56*(t-5)+ 50*(t-6)
215= 56t - 280 + 50t - 300
215 + 280 + 300 = 56t + 50t
795 = 106t
7.21...= t

So, they will be passing each other at about 7:10 pm?

Please advise...

Ang-sui

 

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Followup To
Question -
Hi Bobby,

will you please take me through the steps to solve the following question?

An airplane leaves LaGuardia Airport at 3 p.m. and flies west at an average rate of 200 m.p.h. At 3:30 p.m., another airplane leaves the airport and flies west at an average rate of 250 m.h.h. At what time does the second plane overtake the first plane?

Thanks and regards,

Ang-sui

Answer -
Hi Ang-sui,

Here is how we solve this problem.  We first want to set up equations that represent the distance each plane traveled with respect to time.  For the first plane, we know that it started at 3:00 and flew at 200mph.  So, at 3:00 pm, the plane had flown 0 miles and at 4 pm the plane had flown 200 miles, etc.  The normal equation for distance is speed times time.  We will offset this by three since he started at 3:00pm.  Here is the equation.

x1 = (t - 3)*200

Notice that at t=3, x1 = 0 and at t=4, x1 = 200 just like we said.  For the second plane we can write

x2 = (t - 3.5)*250

Notice that we use 3.5 here because this plane left at half past three.  Now we want to know when (t) the second plane over takes the first one, or when does x2 = x1.  Let's put those two equations together and solve for t.

(t-3)*200 = (t-3.5)*250
200t - 600 = 250t - 875
275 = 50t
t = 5.5

So, the second plane will overtake the first plane at 5:30 pm.  Let me know if you have questions.

Bobby
Answer -
Hi,

Here is the way I solve this one.  You have to set a single reference.  Say that jill starts at a distance x = 0 and billy starts at a distance x = 215.  Then the equations for each are:
x1 = (t-5)*56
x2 = 215 - (t-6)*200

When I set these equal to each other, I get t = 6.62.

This corresponds to about 6:37pm.

Bobby
Answer -
Hi,

I'm not sure how the answer could be 7:30.  Here is the problem.

**At 5 p.m., Jill leaves New York City and travels toward Boston at an average rate of 56 m.p.h. At 6 p.m., Billy leaves Boston and travels toward New York, on the same road as Jill, at an average rate of 200 m.p.h. If they were initially 215 miles apart, at what time do Billy and Jill pass each other?**

Please read it carefully so I know that everything is right.  At 7:30, billy would have been traveling for 1.5 hrs.  That means he would have traveled 300 miles which is farther that they were apart to begin with.  Please check the problem facts, and then the answer.

Bobby

Answer
Small error in your math.  It should be 300 instead of -300 below because a negative times a negative equals a positive.  You almost got it.

So, it must be:
x1 = (t-5)*56
x2 = 215 - (t-6)*50
then
56t - 280 = 215 - 50t + 300

56t + 50t = 280 + 215 + 300
   
106t = 795
t = 7.5 or 7:30 pm.

Bobby