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Algebraic Math Problem

  • Thread starter eNathan
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  • #1
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Hey fellow PhysicForumer's!

I just started school recently and forgot some of my algeberic skills (I will need them!). Can somenody please explain how to derive an answer to this question.

A bike leaves townat 15 miles per hour at 8:00 AM. A car leaves at 11:00 AM traveling in the same direction at 60 miles per hour. How many minutes after 11:00 will the car overtake the bike?

Thanks in advanved!
 

Answers and Replies

  • #2
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derive an equation for both displacements using a single value of t (time) that is adjusted in such a way in one of the euqtions such that it accounts for the difference in the times at which the vehicles departed, then equate these two euqations and solve for t.
 
  • #3
Well, first you use the common physics formula:
s1 + v1t = v2t
s1 is how far ahead the first bike is at 11AM (Simple calculation) And v1 is its velocity while v2 is the car's velocity...
Then just solve for t. (If I'm wrong here someone please tell me)
 
  • #4
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it will take one hour
 
  • #5
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Using Pseudo Statistic's method, I derived
[tex]45 + 15t = 60t[/tex] OR
[tex]45 + 15 = \frac {60t} {t}[/tex] OR
[tex]60 = 60[/tex] Remember that the 2 t's will cancel out at the right of the equals sign. So we are left with 60=60, what does this mean? 60 minutes? 60 miles per hour? But 60=60 makes no sense because you could also derive
[tex] \frac {45 + 15} {60} = \frac {t} {t}[/tex] OR
[tex]1 = 1[/tex]

So how would you solve for t?

/me = confused but catching on
 
Last edited:
  • #6
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A bike leaves townat 15 miles per hour at 8:00 AM. A car leaves at 11:00 AM traveling in the same direction at 60 miles per hour. How many minutes after 11:00 will the car overtake the bike?
at t = 8am bike leaves at 15mph, our basic unit of time will be the hour, so for each hour t, the bike will have the equation

[tex] x_{bike} = 15t [/tex]

The car will leave 3 hours later and has the velocity equation

[tex] x_{car} = 60t [/tex]

The only thing left to do is to compensate for the fact that the bike left 3 hours earlier.. well in that 3 hours, the bike travelled 15miles/hour * 3 hours = 45 miles, so when the car starts moving the bike already has 45 miles under its belt, so now takign 11am as our reference point, we can describe the bike's position equation as

[tex] x_{bike} = 15t + 45 [/tex]

and the car as

[tex] x_{car} = 60t [/tex]

Now, the two cars will be at the same point when their position equations are equal, or

[tex] 60t = 45 + 15t [/tex]

Subtract like terms and solve for t. Your final answer will have units of hours in the form of t = xxx hours and will correspond to the number of hours after 11am.
 
  • #7
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Well put whozum.
[tex]60t = 45 + 15t[/tex]
[tex]60 = \frac {45 + 15t} {t}[/tex]
[tex]60 = 60[/tex]
[tex]1 = 1[/tex]

So 60 minutes (1 hour) after 1:00 the car will overtake the bike, makes perfect sense :D

Thanks a ton!
 
  • #8
Integral
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No, you still have not got the algebra correct. You cannot, nor do you want to, eliminate the variable by division.

you must isolate it. Try this


at t=0 the car leaves, the bike has traveled t+3 hours:
15(t+3) = 60t
distribute the 15
15t + 45 = 60t
now eliminate the t from one side
(15t + 45) -15t = 60t -15t

Can you continue?
 
  • #9
379
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eNathan said:
Using Pseudo Statistic's method, I derived
[tex]45 + 15t = 60t[/tex] OR
[tex]45 + 15 = \frac {60t} {t}[/tex] OR
[tex]60 = 60[/tex] Remember that the 2 t's will cancel out at the right of the equals sign. So we are left with 60=60, what does this mean? 60 minutes? 60 miles per hour? But 60=60 makes no sense because you could also derive
[tex] \frac {45 + 15} {60} = \frac {t} {t}[/tex] OR
[tex]1 = 1[/tex]

So how would you solve for t?

/me = confused but catching on
you can not divide by t because t is not a common factor on the left. you must subtract 15t from the left and the right and then you can isolate t.
 
  • #10
Yay, so after all my method was right. :D
Wow I'm soo smart. ;) *sarcasm*
Your only mistake was you would have gotten 45/t + all the others, as 45 divided by t doesn't yield 45.
Anyway, the logic behind the problem is that you're supposed to find the point where both "vehicles" are at the same position... and yeah.
 
  • #11
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LOL I was still wrong. I get it now. You just move the +15t on the left to the right, and derive 45 = 45t, cancel out the 45, and derive 1 = t.

Thanks agian
~eNathan
 
  • #12
Integral
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Science Advisor
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You have one subtle but very important error remaining.

To be completely correct you must show the units of your answer.

T=1 What?
Seconds? Minutes?
You should be able to determine from the problem what the units need to be. If you carry units on all of your numbers it provides both a check and a guide to the correct solution.
 
  • #13
352
1
Integral said:
You have one subtle but very important error remaining.

To be completely correct you must show the units of your answer.

T=1 What?
Seconds? Minutes?
You should be able to determine from the problem what the units need to be. If you carry units on all of your numbers it provides both a check and a guide to the correct solution.
The Units would be in hours because the units of speed is in miles per hour. Im not saying it as if you arnt aware of that lol. Actually, in the education system which I work on (a computer based system :smile: ), the software only takes numbers and floats, it doesn't accept units of measurement if your actaul answer. But it tells you what units to use in the question.
 
  • #14
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o.k here we go...

60 (t-3) = 15t

60t - 180 = 15t

60t - 15t = 180

45t = 180

t = 180/45

t= 4.0

The intercept point will be 4 hours after the bicycle left; or 12:00. So, it's 1 hour after

11:00.
 

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