Velocity

Velocity is a fancy way of saying speed, or the distance an object has travelled in a certain amount of time. In physics, it is measured in meters per second, or ms^-1.

Velocity is speed with a vectorIf you don't know what a vector is, check out the Displacement section.. Speed is only the measurement.

An example of velocity is 4ms^-1 to the right.
An example of speed is 4ms^-1. See the difference?

You'll see a lot more vector business before you reach the end (Unless of course you skip straight to the bottom, but then you'll miss out on the velocity work in between, and we don't want that).

The formulas for speed and velocity are almost exactly the same; however they only calculate the average.
To calculate the average speed, take the distance travelled and divide it by the time taken.

However, if you want to calculate the average velocity, you take the displacement and divide it by time like so:

Let’s have an example of a straight line velocity question, shown in the flash animation below.

So the marble rolls 15m in 3s. What was its average velocity or speed?

We know that distance = 15m, and that time = 3s. If we put these into the formula we get:

v = 15÷3
v = 5ms^-1This is a speed measurement, because we haven't mentioned what direction the marble traveled in. (1 s.f.Don't know what this is? Significant figures are explained in the Before We Begin section.)

Sometimes, the measurements taken won’t be in the correct SI unitsA list of these can be found in the Before We Begin section..

Let’s say a car drove along a ·75km stretch of road in 1 minute. Is the road more likely to be a town road or a highway?

Firstly, we need to change the measurements to their proper SI units.

·75km = 750m
1 minute = 60 seconds

Now that we’ve done that, we can put them back into the equation.

v = 750÷60
v = 12·5ms^-1

We’re not quite finished, though. We need to know how many kmh^-1This is a fancy way of writing kilometers per hour, e.g. the highway speed limit for a car is 100kmh^-1. that is.

To convert it, we can multiply and divide the answer in bits.

1m = ·001km
12·5÷1000 = 0·0125kms^-1

It’s still not finished, because we need to convert the seconds into hours.

1 hour = 3600 seconds
0·0125×3600 = 45kmh^-1 (2 s.f.)

That’s the answer, so it was more likely a town road.

Ok, so we know how to calculate velocity if the object travels in a straight line, but what if it was a car travelling around a town? Let’s take this example.

We can easily figure out the distance travelled, but take a moment to think about whether that’s the measurement we’re looking for. We actually need the displacement to find out the average velocity.

So what is the average velocity if this trip was taken in 4 minutes (240 seconds)?

We know the distance travelled is 3km, but that’s not the displacement.

We’re going to assume you know that the displacement is 1200mIf you didn't know this, it is worked out on the Displacement page., so we can go straight on to the formula.

v = d÷t
v = 1200÷240
v = 5·00ms^-1 (3 s.f.)

BUT WAIT! There is one final detail we’re missing. The direction.

We know that the car ends up to the right of where it started, so the velocity must be 5·00ms^-1 to the right.

The next example takes us around town for a second time. For more on this example, look at the Displacement section, Example 4. The trip path is shown below.

We are going to save you the trouble of figuring out the displacement1300m and the bearing157°, but if you want to know how to figure it out, look at Example 4 on the Displacement page.

If this trip took 2 minutes (120 seconds), so what is the velocity?

Remember, to find the velocity we use the displacement, which is 1300m.

To find the velocity we use the formula as before.

v = d÷t
v = 1300÷120
v = 10·8

But we need to include the direction, so we write "10·8ms^-1 at 157° (3 s.f.)".

Trigonometry can also be used in the other direction. Say for instance we have a remote-controlled car that has two sets of wheels, allowing it to travel in two directions at once. At one stage, is travelling 9ms^-1 upwards and 5ms^-1 to the right at the same time. What is the angle and overall velocity?

To the right is a diagram for the above question. To get started, lets find the overall velocity. This is done with the Pythagorean Theorem (a²+b² = c²).

So to get value v we turn the formula into this:

a²+b² = v²
9²+5² = v²
v² = 81+25
v² = 106
v = 10·3ms^-1 (3 s.f.)

For the angle we bring out the tool of SOHCAHTOA again; and yet again we will use the "tan" function.

angle = tan^-1(9÷5)
angle = tan^-1(1·8)
angle = 60·9° (3 s.f.)

So our final answer is...

10·3ms^-1 at 60·9° (3 s.f.)

Note: If you have more than two vectors, string them all together into a chain, and work out the displacement the same way. Too many vectors can be troublesome though, which is why we have only used two for the examples.

Once you’ve got used to dealing with velocity in straight lines and paths, you can move on to Acceleration or Relative Velocity.