Runner and a bird heading towards finish line

  • Thread starter anglum
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In summary: just think of it as two runners... they leave the start position at the same time... one finishes the race 51.4 s after the other... use t as time taken by one wave... what's the time taken by the other wave... what's the distance in terms of...if the earthquake occurs at 4:23pm and the seismograph records the arrival of the transverse waves at 5:14pm then the earthquake is 4.93 km away from the seismograph
  • #1
anglum
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Homework Statement


A runner is jogging at a steady velocity of 3.4 km/h. When the runner is 4.1km from the finish line, a bird begins flying from the runner to the finish line at velocity 10.2km/h. When the bird reaches the finish line it turns around and flies back to the runner. Assume the birds length equals zero... How far does the bird travel?

The Attempt at a Solution



i solved that it would take the bird .4019h to get to the finish line from that point...

at that time the jogger is 2.733 km from the finish

what are my next steps to calculate how far the bird goes to reach the runner... since the runner would be moving as the bird moves back towards it??
 
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  • #2
since hte bird travels 3 times faster than the jogger and we know at the point the bird is at the finish line... that the distance away is 2.733333 then can we divide that by 4 and then multiply by 3 to get the distance tehy are at when they intersect again? since if the bird travels 3 times faster then the distance would be 3 times further it travels at this point and would be 2.049975 of the possible 2.7333 distance after reaching the finish line

therefore i add 2.049975 and 4.1 km to get my final answer for how far the bird traveled?
 
  • #3
so the distance the bird traveled was 6.149975 km

now for part two of the problem PLEASE HELP

After the first encounter the bird then turns and heads back to the finish line, turns around again and heads back to the runner. then repeats that until the runner reaches the finish line... how far does the bird travel including the 6.149975 distance from the first part of the answer. answer in units of km

please help as i am stuck on this one... is there a formula to use
 
  • #4
You've calculated that when the bird hits the finish line... the runner is 2.733 km away...

Suppose t seconds beyond that point the runner and the bird meet... what is the distance the runner traveled in terms of t... what is the distance the bird ran in terms of t... what is the sum...


EDIT: did you finish the first part? For the second part... they want the total distance from when the runner was at 4.1km... from that point... how long does it take the runner to hit the finish line?
 
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  • #5
i have done that,... do i just keep repeating those steps over and over again until the runner is 0km away
?

seems like a lot of work to get that answer
 
  • #6
anglum said:
i have done that,... do i just keep repeating those steps over and over again until the runner is 0km away
?

seems like a lot of work to get that answer

No... when from when the runner is 4.1km away... how long does it take the runner to hit the finnish line?
 
  • #7
if the runner is 4.1 km away... it takes him 1.205882353 hours to cross the finish line...

so if that is the total time it takes him to cross the line do i just multiply that by the velocity of the bird to get its total distance flown?
 
  • #8
anglum said:
if the runner is 4.1 km away... it takes him 1.205882353 hours to cross the finish line...

so if that is the total time it takes him to cross the line do i just multiply that by the velocity of the bird to get its total distance flown?

yup.
 
  • #9
dude u are a genius at this stuff

i have 2 more iam struggling with... should i post them in here or in a diff thread... mind helpin me out a lil?
 
  • #10
anglum said:
dude u are a genius at this stuff

i have 2 more iam struggling with... should i post them in here or in a diff thread... mind helpin me out a lil?

you can post here.
 
  • #11
the velocity of the transverse waves produced by an earthquake is 4.93 km/s while the longitudinal waves are 8.8247 km/s

a seismograph records teh arrival of the transverse waves 51.4s after that of the longitudinal waves

how far away was the earthquake

i calculated the longitudinal waves travel 1.79 times faster than the transverse waves
 
  • #12
does knowing how many times faster one travels than the other help me out at all?
 
  • #13
anglum said:
the velocity of the transverse waves produced by an earthquake is 4.93 km/s while the longitudinal waves are 8.8247 km/s

a seismograph records teh arrival of the transverse waves 51.4s after that of the longitudinal waves

how far away was the earthquake

i calculated the longitudinal waves travel 1.79 times faster than the transverse waves

just think of it as two runners... they leave the start position at the same time... one finishes the race 51.4 s after the other... use t as time taken by one wave... what's the time taken by the other wave... what's the distance in terms of t?
 
  • #14
im not sure how to do this one without knowing the distance... it is really jumbled in my head

so runner 1 goes 8.8247 km/s

runner 2 goes 4.93 km/s
and 51.4s slower

i feel like i need one other number to make the next step
 
  • #15
anglum said:
im not sure how to do this one without knowing the distance... it is really jumbled in my head

so runner 1 goes 8.8247 km/s

runner 2 goes 4.93 km/s
and 51.4s slower

i feel like i need one other number to make the next step

runner 1 takes t seconds to finish the race. What distance does it travel in terms of t? If runner 1 takes t seconds how many seconds does runner 2 take? What distance does does runner 2 travel in terms of t?
 
  • #16
ok so runner 1 goes 8.8247 km/s and he takes t seconds to finish therefore his distance is 8.8247/t

runner 2 goes 4.93km/s and he takes t+51.4 seconds to finish nad his disctance is 4.93/t+51.4

correct?

so where do i go from there?
 
  • #17
anglum said:
ok so runner 1 goes 8.8247 km/s and he takes t seconds to finish therefore his distance is 8.8247/t

runner 2 goes 4.93km/s and he takes t+51.4 seconds to finish nad his disctance is 4.93/t+51.4

correct?

so where do i go from there?

you mean multiply... not divide... multiply the velocity and time... can you relate the two distances? then you can solve for t.
 
  • #18
what do mean by relate the 2 distances

if runner 1 ----- d = 8.8247 * t

if runner 2 ------ d = 4.93 * (t+51.4)

how would i relate those 2 now?
 
  • #19
anglum said:
what do mean by relate the 2 distances

if runner 1 ----- d = 8.8247 * t

if runner 2 ------ d = 4.93 * (t+51.4)

how would i relate those 2 now?

Both runners are running the same race. :wink:
 
  • #20
so 4.93 * ( t+51.4) = 8.8247 * t

can i then deduce 4.93t + 253.402 = 8.8247t
 
  • #21
anglum said:
so 4.93 * ( t+51.4) = 8.8247 * t

can i then deduce 4.93t + 253.402 = 8.8247t

yup solve for t... then get the distance of the race.
 
  • #22
so i then get 253.402 = 8.8247t - 4.93t

253.402 = 3.8947t

65.06329114 = t

is that correct ?
 
  • #23
yup i got it right... ok learning can you help me one last time please?
 
  • #24
anglum said:
so i then get 253.402 = 8.8247t - 4.93t

253.402 = 3.8947t

65.06329114 = t

is that correct ?

yes that's right. now you can get the distance.
 
  • #25
how long does it take car traveling 68.2km/h to become even with a car that is traveling in another lane at 56.2km/h if the front bumpers are initially 131 m aparti have tried this so far

do i need to convert the 131 m to . 131 km??

d = 56.2km/h * t

and

d +131 = 68.2 km/h * t

and i know that the time is equal for both

i get d/56.2 = d+131/68.2

correct??

how and i solve for d i get the distance of the race?
 
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  • #26
anglum said:
how long does it take car traveling 68.2km/h to become even with a car that is traveling in another lane at 56.2km/h if the front bumpers are initially 131 m apart

i have tried this so far

d = 56.2km/h * t

and

d +131 = 68.2 km/h * t

and i know that the time is equal for both

i get d/56.2 = d+131/68.2

correct??

how and i solve for d i get the distance of the race?

yes:
d/56.2 = (d+131)/68.2

and solve for d. Did you get the previous problem?
 
  • #27
yes i got the previous problem

should the 131 actually be .131 km or does it not matter?

this is a horrible question and u can tell its gettin late

but when its setup like that how do i solve for d?
 
  • #28
anglum said:
yes i got the previous problem

should the 131 actually be .131 km or does it not matter?

this is a horrible question and u can tell its gettin late

but when its setup like that how do i solve for d?

yeah, just evaluate out 1/56.2, 1/68.2 and 131/68.2 in your calculator...

ie your equation is:

(1/56.2)d = (1/68.2)d + (0.131/68.2)
 
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  • #29
does the 131 m need to be converted to .131 km before i do that

im not sure what u mean evaluate out all 3 of thos ein my calculator
 
  • #30
anglum said:
does the 131 m need to be converted to .131 km before i do that

im not sure what u mean evaluate out all 3 of thos ein my calculator

yeah rewrite as .131km sorry... have a look at the way I rewrote the equation... you want those numbers in parentheses...
 
  • #31
so i then get

.0177935943D=.0146627566D + .0019208211

i then get .0031308377D = .0019208211

D = .6135166636 km

that is the D then i solve for t of the equation

t = .6135166636/68.2

t = .008 hours? that doesn't seem right to me
 
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  • #32
anglum said:
since the cars travel at km/s and the distance between them is 131 m... doesn't that have to be .131 km in the equation to get a true correct answer?

yeah, that's right sorry.
 
  • #33
anglum said:
so i then get

.0177935943D=.0146627566D + .0019208211

i then get .0031308377D = .0019208211

D = .6135166636 km

that is the D then i solve for t of the equation

t = .6135166636/68.2

t = .008 hours? that doesn't seem right to me
i then take .008 hours and convert to minutes which is .5397507304 mins then convert again to seconds by multiplying by 60 which gives me a time in seconds of 32.385

? is that the answer that you got for this problem? because i get that as being wrong when entered on the classroom online

this was the original info

how long does it take car traveling 68.2km/h to become even with a car that is traveling in another lane at 56.2km/h if the front bumpers are initially .131 km apart
 
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  • #34
something in my calculations must be wrong or else the methods i used to solve were not correct
 
  • #35
learning have u done this problem and gotten an answer in seconds for the time it would take for them to be even and directly next to each other
 

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