DiscoverHover CURRICULUM GUIDE #2
VELOCITY AND ACCELERATION
© 2004 World Hovercraft Organization
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Before beginning to discuss how hovercraft and other objects move, we need a way to measure motion. Terms like speed, velocity, and acceleration are all familiar, but let’s investigate exactly how they are related and how they are measured.

Speed is a change in an object’s position with respect to time. This can be found by measuring how far an object travels in a certain amount of time. For example, if a hovercraft traveling at a constant speed goes 100 miles [161 km] in two hours, its speed was 50 mph (miles per hour) [80.5 km/h]. Just divide how far the hovercraft went by how long it took to get there. Often the term ‘velocity’ is used instead of speed, but these two terms are a little different. Velocity is a measure of both speed and direction of an object. Imagine one hovercraft traveling 60 mph [96.6 km/h] north on a river, and another hovercraft traveling 60 mph [96.6 km/h] south on a river. These two hovercraft would have the same speed, but different velocities because they are traveling in different directions. As you can see, velocity is a more detailed description of how fast something is traveling. An equation for calculating velocity is given below.

Velocity = change in position
change in time

Acceleration measures the change in velocity with respect to time. It’s pretty clear that speeding up and slowing down are both forms of acceleration, but remember that acceleration is change in velocity, not necessarily speed. When a hovercraft turns, even if its speed isn’t changing, it’s accelerating because its direction, and therefore velocity, is changing.

Acceleration = change in velocity
change in time

In our world, objects are constantly accelerating and decelerating, and this can become quite difficult to test with a model. If we only look at cases in which the acceleration is constant, it becomes much easier to model and to calculate. In fact, we can model any motion along a straight line using only 4 formulas as long as we assume acceleration is always constant.

Formulas for Motion when Acceleration isn’t Changing

v = v0 + a·t

x = x0 + ½ (v + v0)t

v² = v0² + 2a(x – x0)

x = x0 + v0t + ½ a·t²

In these formulas, ‘x’ refers to the position of the object, ‘v’ refers to its velocity, ‘a’ gives its acceleration, and ‘t’ gives the time at which these variables are measured. You can see that some of the variables have little zeros next to them. For example, ‘x0’ is pronounced “ x naught” (sounds like “not”). These are used to represent initial conditions, so ‘x0’ represents the position of the object initially (at time t =0), while ‘x’ represents the position of the object at time, ‘t’. We’ll do a few examples to see how these formulas work.

Example 1: (Using Imperial units)
Starting from a stop, a hovercraft accelerates at 5 feet per second squared. How fast will the hovercraft be traveling when it reaches 500 ft from where it started?

Solution:
Let’s begin by listing the variables that we’re given and the ones we need to find. We are given the acceleration, and since the hovercraft starts from a stop, we know that the initial velocity is zero. Since we know the hovercraft travels 500 ft, we can say the initial position is 0 ft and the final position is 500 ft. We need to solve for the final velocity.

a = 5 ft/s²   v0 = 0 ft/s    x0 = 0 ft    x = 500 ft    v = ? ft/s

Now we need to find one of the formulas just mentioned that has the one variable we’re looking for as well as variables that we have values for. It looks like formula (3) has everything we need and nothing we don’t. Now we can put in all the values and calculate the final velocity.

v² = v0² + 2a(x – x0)
v² = (0 ft/s)² + 2 (5 ft/s²) (500 ft – 0 ft)
v² = 2 (5 ft/s²) (500 ft)
v² = 5000 ft²/s²
v = 70.7 ft/s


500 ft from where it started, the hovercraft will be traveling 70.7 ft/s [48.2 mph].


Example 2: (Using Imperial units)
The same hovercraft accelerates from a stop at 5 ft/s². How far will it travel and how fast will it be going after 10 s?

Solution:
Again we’ll start by listing what variables we know and which ones we need to find.

a = 5 ft/s²    v0 = 0 ft/s    x0 = 0 ft   t = 10 s   x = ? ft    v = ? ft/s

If we start by just trying to find ‘x’, it looks like equation (4) is the one to use.

x = x0 + v0t + ½ a·t²
x = (0 ft) + (0 ft/s) (10 s) + ½ (5 ft/s²) (10 s)²
x = ½ (5 ft/s²) (100 s²)
x = 250 ft


We know that the hovercraft will be 250 ft away after 10 s, now we need to find how fast it will be going when it gets there. We can find this with equation (1).

v = v0 + a·t
v = (0 ft/s) + (5 ft/s²) (10 s)
v = 50 ft/s

After 10 s, this hovercraft will be 250 ft away from where it started, traveling 50 ft/s.

The first two examples we calculated using Imperial units. Now we’ll try one using System International(SI) units.

Example 3: (Using SI units)
A hovercraft drives past a bush traveling at 15 m/s and continues traveling towards a tree standing 200 m away from the bush. The hovercraft passes the tree 10 s after it passed the bush. Assuming the hovercraft maintained a constant acceleration, how fast was it traveling when it passed the tree?

Solution:
This problem is a little harder, but if we stick to the same strategy we can figure it out.

v0 = 15 m/s   x0 = 0 m    x = 200 m   t = 10 s   v = ? m/s

For this one we’re going to use equation (2), but ‘v’ is buried inside the equation so we have to get it out.

x = x0 + ½ (v + v0)t
(200 m) = (0 m) + ½ (v + 15 m/s) (10 s)
(200 m) = (5 s) (v + 15 m/s)
(40 m/s) = (v + 15 m/s)
v = 25 m/s

Our calculation shows that the speed of the hovercraft increases from 15 m/s to 25 m/s by the time it reaches the tree.

 

As you can see in the examples, the trick to figuring out which formula to use for these problems is to list all the variables you have values for and the variables you want to find. Pick out a formula that contains only one unknown variable.