wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Rectilinear problem
        
(Message started by: Larissa_Preedy on Jun 10^th, 2005, 11:38pm)

Title: Rectilinear problem
Post by Larissa_Preedy on Jun 10^th, 2005, 11:38pm

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!

Title: Re: Rectilinear problem
Post by Sir Col on Jun 11^th, 2005, 1:11am

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

Title: Re: Rectilinear problem
Post by Larissa_Preedy on Jun 11^th, 2005, 3:33am

yay it worked!

wat was wrong with my method thougH? :(

or was i totally on the wrong trac?

Title: Re: Rectilinear problem
Post by Sir Col on Jun 11^th, 2005, 7:50am

on 06/11/05 at 03:33:45, Larissa_Preedy wrote:

wat was wrong with my method thougH? :(

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.