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Topic: Rectilinear problem (Read 1463 times) |
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Larissa_Preedy
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Rectilinear problem
« on: Jun 10th, 2005, 11:38pm » |
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hey guys, this one is really important, and most likely to be the last one i need help with.
Given a = 4v/100
where a is in m/s2 and v is in m/s
and given that when t =1s, v = 5m/s
what is the velocity by the time t = 10s
This is my working
a = 6v/100 and V(3) = 8
v = e^(6*t/100) + c
v(3) = e^((6*3)/100) + c = 8
c = 6.8028
v(t) = e^(6*t/100) + 6.8028
v(8) = e^((6*8)/100)
i'm totally stuck, can't see where i'm wrong :(
can someone help me urgently.. thanks!
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| « Last Edit: Jun 10th, 2005, 11:45pm by Larissa_Preedy » |
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Sir Col
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impudens simia et macrologus profundus fabulae
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Re: Rectilinear problem
« Reply #1 on: Jun 11th, 2005, 1:11am » |
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I'd do it by first separating the variables:
a = dv/dt = 4v/100 = v/25
1/v dv = 1/25 dt
Integrate both sides:
ln(kv) = t/25
kv = e^(t/25)
When v = 5, t = 1
5k = e^(1/25)
k = e^(1/25)/5
v*e^(1/25)/5 = e^(t/25)
v = 5*e^(t/25)/e^(1/25)
Hence v = 5*e^((t-1)/25)
So when t = 10, v = 5*e^(9/25) ~= 7.17 m/s
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Larissa_Preedy
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Re: Rectilinear problem
« Reply #2 on: Jun 11th, 2005, 3:33am » |
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yay it worked!
wat was wrong with my method thougH? 
or was i totally on the wrong trac?
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Sir Col
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Re: Rectilinear problem
« Reply #3 on: Jun 11th, 2005, 7:50am » |
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on Jun 11th, 2005, 3:33am, Larissa_Preedy wrote:| wat was wrong with my method thougH? |
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You seem to have changed all the numbers in the problem in your solution: a=4v/100 became a=6v/100; v=5 became v=3; and t=10 became t=8?
Also you made an error after you integrated.
From 1/v dv = 4/100 dt, you got ln(v) = 4t/100 + c, but then you anti-logged both sides and didn't handle the c properly.
You should have got, v = e^(4t/100+c), not e^(4t/100)+c.
From this you would get, 5 = e^(4/100+c), c = ln(5)–4/100 ~= 1.569
So v ~= e^(40/100+1.569) ~= 7.17.
I much prefer to place my constant in with the logarithm ln(kv), as it generally allows you to obtain a more concise exact formula.
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