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Old Jan 28th 2009, 11:44 PM   #1
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Kinematics Help

Hi,
I need a formula for this: a man runs to the airport check-in. The distance to the airport check-in is 1150 meters. His weight is 80kgs. His baggage is 10kgs, which he carries by hand. He takes 6 minutes to get to the counter.
Question:
What is the formular to calculate how long he will take to get to the counter if his baggage was 20kgs.

Thanks

Last edited by janvdl; Mar 20th 2009 at 01:01 AM. Reason: More descriptive title given.
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Old Mar 19th 2009, 02:06 AM   #2
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From s = ut + 1/2 a t^2 calculate his acceleration a with u = 0 m/sec.a works out to be = 0.0177 m/sec^2.
The max force he can exert can be found by
F = ma = (80+10) x 0.0177. = 1.597
In the second case also this is the max force he can exert. So,
1.597 = (80+20) a1 . So a1 = 0.01597 m/sec.

Use s = (1/2) x a1 x t1^2.

t1 = 6.325 min or 6min,19.5 sec.

We cant treat this as a problem on friction.
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Old Mar 20th 2009, 12:07 AM   #3
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Originally Posted by physicsquest View Post
From s = ut + 1/2 a t^2 calculate his acceleration a with u = 0 m/sec.a works out to be = 0.0177 m/sec^2.
The max force he can exert can be found by
F = ma = (80+10) x 0.0177. = 1.597
In the second case also this is the max force he can exert. So,
1.597 = (80+20) a1 . So a1 = 0.01597 m/sec.

Use s = (1/2) x a1 x t1^2.

t1 = 6.325 min or 6min,19.5 sec.

We cant treat this as a problem on friction.
Hi,
Thanks for the result,
Please can you give me the detailed steps in getting to the results. I am confused.

Thanks in advance for your kind assistance.

Regards
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Old Mar 20th 2009, 01:24 AM   #4
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I think i have outlined most of the steps and worked it out.

Only Newton's law F = m x a and the equations of motion have been used.
The idea is that the man is capable of exerting only the same force in both the cases, but he carries a heavier load in the second case and so takes longer.
Please specify where exactly the confusion lies.
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Old Mar 20th 2009, 06:41 AM   #5
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Originally Posted by physicsquest View Post
I think i have outlined most of the steps and worked it out.

Only Newton's law F = m x a and the equations of motion have been used.
The idea is that the man is capable of exerting only the same force in both the cases, but he carries a heavier load in the second case and so takes longer.
Please specify where exactly the confusion lies.
Hi,
I need the workings in the formula step by step on how to get to the answer of 6.325.
Example what is t1^2 etc..
Note: I am a not math literate...as yet. Hope you dont mind.

thanks
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Old Mar 20th 2009, 11:47 AM   #6
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t1^2 is t1 squared or t1 x t1. This notation is being used here as it is not directly possible to show something raised to a power here(without using word)
And speaking of the math, someone who is expected to solve this is also supposed to be exposed to qudratic equations etc.
Please look up the equations of motion and try again.
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