# 1. Velocity (*s-t*) Graphs

### Example 1

This general graph represents the motion of a body travelling at **constant velocity**. The graph is **linear** (that is, a straight line).

Recall that linear equations have the general form

y=mx(wheremis a constant andxis a variable).

The number *m* is called the slope of the line (the vertical rise over the horizontal run).

In the above graph, we have the function:

displacement = velocity × time

or

s=v×t

Velocity is constant and time is a variable.

**NOTE: ** We use the variable "*s*" for displacement. Be careful not to confuse it with "speed"!

We note that the graph passes through `(0,0)` and has slope *v*. The slope of the line tells us the velocity. We can also write the velocity using delta notation:

`v=(Deltas)/(Deltat)`

which means

"change in displacement over change in time".

If we have a high velocity, the graph has a steep slope. If we have a low velocity the graph has a shallow slope (assuming the vertical and horizontal scale of each graph is the same).

### Example 2

A marathon runner runs at a constant `12` km/h.

a. Express her displacement travelled as a function of time.

b. Graph the motion for `0 ≤ t ≤ 4\ "h"`

Answer

a. `s = 12t`*,* for displacement *s* and time
*t.*

b. Graph of `s = 12t`*.*

We stop the graph at (4, 48).

Get the Daily Math Tweet!

IntMath on Twitter

## Displacement-time Graphs

### Example 3

This is the graph of a journey by sports car:

a. What is the velocity for each stage of the journey?

b. What is the average (**mean**) velocity for the whole
journey?

Answer

a. This table outlines the stages of the journey.

12:00 to 12:30 | Travelled `50` km in `30` minutes, so `100` km/h |

12:30 to 13:00 | Stopped |

13:00 to 14:00 | Travelled `50` km in `60` minutes, so `50` km/h |

14:00 to 14:30 | Stopped |

14:30 to 15:00 | Travelled `100` km back towards the starting point in `30` minutes, so `-200` km/h |

b. Even though the whole journey was `200\ "km"` (`100\ "km"` out and `100\ "km"` back)
in `3` hours, the **displacement** for the journey (the distance from the starting point) is `0\ "km"`.

So the average **velocity** is `0\ "km/h"`.

On the other hand, the average **speed** was `"distance travelled"/"time taken" = 200/3 = 66.7\ "km/h"`.

In summary,

`"ave velocity" = "displacement"/"time"`

`"ave speed" = "distance"/"time"`

### Example 4

A particle in a magnetic field moves as follows:

Find the velocity for each part of the motion.

Answer

For `t = 0` to `2`: `v = (Deltas)/(Deltat)=4/2=2\ text(ms)^-1`

(The particle is moving away from the origin, at a constant velocity of `2\ text(ms)^-1.`)

For `t = 2` to `7`: `v = (Deltas)/(Deltat)=0/5=0\ text(ms)^-1`

(The particle stops for `5` seconds.)

For `t = 7` to `8`: `v = (Deltas)/(Deltat)=(-4)/2=-4\ text(ms)^-1`

(The velocity is negative for the last stage, since the particle is going in the opposite direction, back towards the origin.)

### Search IntMath, blog and Forum

### Online Math Solver

This math solver can solve a wide range of math problems.

Go to: Online algebra solver

### Math Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand math lessons on DVD. See samples before you commit.

More info: Math videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!