This is OK. Now at time t=0, x=0, so: k=400 ln(2).Originally Posted by kingkaisai2
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
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!
Hello, kingkaisai2!
A girl lives 500 metres from school.
She sets out walking at 2m/s, but when she has walked a distance of metres
her speed has dropped to 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: .
Multiply by 400: .
Integrate: .
. . When
The equation becomes: .
Divide by
Then we have: .
. . Then: . **
When , we have: .
Then: .
Therefore: .
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**
In order to answer the question, I did not have to solve for
I did it to show how it is done (in case anyone was interested).