Math Is Fun Forum

rscott

Member

Registered: 2008-10-11

Posts: 2

how to: Calculate a point on a line after X minutes?

I'm making a game - I need to know the equation to use to calculate the x,y of where an object is located after a given amount of time. Here's what I know:

1- The object is traveling in a straight line at the same rate of travel
2- The object's start location (x,y)
3- The objects rate of travel
4- The time the object has been traveling and the time the object is supposed to arrive
5- The object's destination (x,y)

Any tips?

Thanks

Ryan

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: how to: Calculate a point on a line after X minutes?

Parametric functions are useful when you want to find individual formulas for an object's x and y coordinates.

o_x: location in the x-axis at time t.
o_y: location in the y-axis at time t.

We know that o_x(0) = x and o_y(0) = y.  We also know that o_x and o_y take on the following form:

o_x(t) = mt + b
o_y(t) = nt + c

Finally, we know the final destination, I will call it (w, z) at time s, when the object reaches it's destination:

o_x(s) = w
o_y(s) = z

So now we solve for m, b, n, and c.

o_x(0) = b = x
o_y(0) = c = y

So b = x and c = y.  Now try solving for n and m, and you will find the objects position at time t expressed by:

(o_x(t), o_y(t))

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

rscott

Member

Registered: 2008-10-11

Posts: 2

Re: how to: Calculate a point on a line after X minutes?

Thanks for the post Ricky...

Wow, my math must be rusty.

Can we try this with some sample numbers?

Object started at (12,30) with a final destination of (-10,10). It will take a total of 10 minutes to get to the destination and has been traveling for 6 minutes so far.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: how to: Calculate a point on a line after X minutes?

Then we have the same equations:

o_x(t) = mt + b
o_y(t) = nt + c

As I said before:

o_x(t) = b = 12
o_y(t) = c = 30

So now we have:

o_x(t) = mt + 12
o_y(t) = nt + 30

Now we know that:

o_x(10) = -10
o_y(10) = 10

So we get:

-10 = o_x(10) = m*10 + 12
10 = o_y(10) = n*10 + 30

And solving this, we find:

m = -22 / 10 = -11/5
n = -20/10 = -2

So that our equations are:

o_x(t) = (-11/5)t + 12
o_y(t) = (-2)t + 30

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."