Differential equation-

**kingkaisai2** · August 28th 2006, 03:57 AM

A girl lives 500 metres from school. She sets out walking at 2m/s, but when she has walked a distance of x metres her speed has dropped to (2-x/400)m/s. How long does she take to get to school.

Can you do the working-

Just if you are interested this is my working for the question-
dx/dt=(2-x/400)
dx/(2-x/400)+k=dt
-400in(2-x/400)+k=t
Is this right?
So after this do I substitute 500 into x or something to find the time- what occurs from here-
I got a really strange answer- it is an incorrect answer. Urgent!

**CaptainBlack** · August 28th 2006, 04:56 AM

Originally Posted by kingkaisai2

[SIZE=5]A girl lives 500 metres from school. She sets out walking at 2m/s, but when she has walked a distance of x metres her speed has dropped to (2-x/400)m/s. How long does she take to get to school.

Can you do the working-

Just if you are interested this is my working for the question-
dx/dt=(2-x/400)
dx/(2-x/400)+k=dt
-400 ln(2-x/400)+k=t

This is OK. Now at time t=0, x=0, so: k=400 ln(2).

Now you want to find the time t_500, when x=500, so plug 500 in for x to get:

t_500=400 ln(2) -400 ln(2-500/400)=392.3s

RonL

**Soroban** · August 28th 2006, 11:20 AM

Hello, kingkaisai2!

A girl lives 500 metres from school.
She sets out walking at 2m/s, but when she has walked a distance of $x$ metres
her speed has dropped to $2-\frac{x}{400}$ m/s.
How long does she take to get to school?

If you hate fractions as much as I do, you'd eliminate them immediately.

We have: . $\frac{dx}{dt}\:=\:2 - \frac{x}{400}$

Multiply by 400: . $400\frac{dx}{dt}\:=\:800 - x\quad\Rightarrow\quad \frac{400\,dx}{800 - x} \:=\:dt$

Integrate: . $\text{-}400\ln|800 - x| \:=\:t + C$

. . When $t=0,\:x=0:\;\;\text{-}400\ln|800 -0| \:=\:0 + C\quad\Rightarrow\quad C$ $\:=\:\text{-}400\ln800$

The equation becomes: . $\text{-}400\ln|800 - x| \:=\:t - 400\ln800$

Divide by $\text{-}400:\;\;\ln|800 - x|\:=\:\text{-}\frac{t}{400} + \ln800$

Then we have: . $800 - x \;=\;e^{\text{-}\frac{t}{400} + \ln800} \;=\;e^{-\frac{t}{400}}\cdot e^{\ln800} \;=\;e^{\text{-}\frac{t}{400}}\cdot800$

. . Then: . $\text{-}x \;=\;\text{-}800 + 800e^{-\frac{t}{400}}\quad\Rightarrow\quad \boxed{x\:=\:800\left(1 - e^{\text{-}\frac{t}{400}}\right)}$ **

When $x = 500$ , we have: . $500 \:=\:800\left(1 - e^{\text{-}\frac{t}{400}}\right)\quad\Rightarrow\quad \frac{5}{8}\:=\:1 - e^{\text{-}\frac{t}{400}}$

Then: . $e^{\text{-}\frac{t}{400}} \:=\:\frac{3}{8}\quad\Rightarrow\quad \text{-}\frac{t}{400} \:=\:\ln\left(\frac{3}{8}\right)$

Therefore: . $t \:=\:\text{-}400\ln\left(\frac{3}{8}\right)\:=\:392.331201 \text{ seconds }\:\approx\: \boxed{6.5\text{ minutes}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

In order to answer the question, I did not have to solve for $x.$

I did it to show how it is done (in case anyone was interested).

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