Printable View
Colin sets off for school, which is 800m from home. His speed is proportional to the distance he still has to go.
Let x meters be the distance that he has gone, and y meters be the distance he still has to go.
a) Sketch the graph of x against t and y against t.
I need help with writing an expression for this word problem in both formats so that I can sketch the graph. I am bad with word problems (Worried)
Hello,
First of all, translate all the information given in the text. Then you may have useless information, but you cannot know at first sight.Quote:
Originally Posted by struck
(No, the fact that the boy's name is Colin is not an important information (Rofl))
Note that y=800-x. Why ?
Let s be Colin's speed.
We know that the speed is proportional to y. This means that there exists a constant k such that $\displaystyle s=k*y$.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let's deal with "x against the time"
Remember that the speed corresponds to the derivative of the distance with respect to the time.
Thus $\displaystyle s=\frac{dx}{dt}$
$\displaystyle s=k*y=k*(800-x)=k'-kx$, where k'=800k.
So you're now with the following equation :
$\displaystyle k'-kx=\frac{dx}{dt}$
I hope you know how to solve a differential equation (Surprised)
Actually, that didn't occur to me. Thanks :DQuote:
Note that y=800-x. Why ?
On another thought, I still don't get it (Doh) ...
Is this the derivative dx / dt?
$\displaystyle
k'-kx=\frac{dx}{dt}
$
P.s. I am not sure why is the k' notation used. Also, I need the $\displaystyle \frac{d^2x}{dt^2}$
bump
k' is just a constant. I wrote it above : k'=800kQuote:
Originally Posted by struck
$\displaystyle \frac{d^2x}{dt^2}=-k \frac{dx}{dt}=-kk'+k^2 x$ (differentiate implicitly and then substitute dx/dt)
