Distance formula

This is just pythagoras' theorem

the difference in the xes is   $$x_2-x_1$$

The difference in the y's is     $$y_2-y_1$$

You want the hyptotenuse (d)

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

The  fundamental distance formula for any problem involving speed and time is:

1. d = s*t (rearranged from s = d/t) and

eg:

speed = 50 kphtime = 2.5 hours

d = 50*2.5 = 125 kilometres.

Here, your distance is 125 kilometres.

2. r - r1 = + v*t (where r is the position vector of an object at a certain point and r1 is the position vector of the same object at t = 0). This method is to be employed when using i and j vector notation.

eg:

velocity = 2i + 3j

t = 5s

r - r1 = 5*(2i + 3j) = 10i + 15j

Then, to find the distance, you must find the modulus* of (r - r1), as you have calculated THE DISPLACEMENT OF THE OBJECT (distance is a scalar quantity, displacement is a vector).

|r - r1| = (10^2 + 15^2)^1/2 = (100 + 225)^1/2 = 325^2 = 18.03 (rounded to 4 significant figures).

So, your distance here is 18.03 units.

*The modulus is to find the magnitude of a vector; you take the value of the vector, square them, add them together, and get their square root. It is the inverse of Pythagoras' theorem.

Hope this was helpful.

P.S: This last answer, the one including vector notation, was posted by me, Dragontmaer (I still haven't got an account here, so this is my name from other forums)

Hi Dragontmaer :)

Why don't you get an accout here.  We would love you to join us and becoming a member is ultra easy :)

Finally i've got an account, but someone already took the name "Dragontmaer", so I took the name "Dragontmaermaths".

Welcome to the forum Dragontmaermaths :))