### Visualizing Motion

In this chapter we will see what the graphs of vector motion look like, as well as see how certain formulae are derived from the graphs.

#### Position V.S. Time

Since position is a vector quantity, position-time graphs contain a direction. Above, the positive position quantities are in the North direction while the negative quantities are in the opposite, South, direction. A position of zero signifies that the object is at the reference point, its starting position.Analysis

- The object is changing its position in the north direction from 0 to 3 seconds, the velocity is uniform.
- The object stays at the same position for about 3 seconds - its
*stopped*for 3 seconds. - From the 6th to the 8th second the object is backtracking, its returning to the starting position ( our reference point ).
- From the 8th second on the object passes the starting position and travels southward from it.

position / time = velocity.

#### Velocity V.S. Time

Analysis- Our data is captured when the object is already moving, it has a constant velocity of 30m/s [N] up to the 3rd second of observation.
- The object decelerates, reaches 0 velocity and then begins travelling in the south direction.
- The object travels at a constant velocity south for 3 seconds
- The object smoothly decelerates and comes to a stop

The slope of the line is the acceleration, but the area under the
the line is the **displacement**!

Let's see how to determine the displacement an object undergoes during a given time:

Determining the displacement of the object in the first 5 ( area 1 ) seconds is simple, the shape is a square and you multiply 20m/s times 5 seconds.

The displacement during those 5 seconds is 100m

**[N]**.

The displacement undergone by the object from 5 seconds to 8 seconds is a bit harder to determine, because the shape under the line is more complex. The area of the square ( area 2 ) is 20m/s times 3 seconds, and represents a displacement of 60m [N]. The are of the triangle ( area 3 ) is 20m/s ( 40 - 20 ) times 3 seconds, divided by two. Thus the area of the triangle ends up reperesenting a displacement of 30m [N].

We can deduct that during the 8 seconds of the object's motion it has undergone a displacement of 100 + 60 + 30 metres northward.

When the line falls blow the 0m/s mark the displacement is happening in the opposite direction. This "negative displacement" should be subtracted from the "positive displacement" to find out what is the resultant displacement and its direction.

#### Uniform Acceleration Equations

Several useful formulae are deductible from the velocity vs time graph of an object's motion. When an object undergoes uniform acceleration its graph looks like a trapezoid, above the object accelerated uniformly up to time*t*.

We can find out the acceleration of the object by finding the slope of the graph. Slope is rise over run:

M = ( Y2 - Y1 ) / ( X2 - X1 )

accleration = ( Vf - Vi ) / ( t - 0 )

a = ( Vf - Vi ) / t

We can model displacement mathematically by formulating a formula for the area of a trapezoid, but using Vf, Vi and t instead of lengths. The formula for a trapezoid is 1/2 ( Y1 + Y2 ) × W where W is the width of the base. Adapting the above formula we get the formula for displacement during uniform acceleration:

1/2 ( Vf + Vi ) × t

#### Acceleration V.S. Time

On the right, however, is the acceleration vs time graph of an object with a varied acceleration. The acceleration apears as a sloped line, indicating that the acceleration is changing. The graph on the right shows an object decelerating, stopping acceleration ( uniform speed ) and then beginning to accelerate in the south direction.

#### Conclusion

The velocity vs time graph is arguably the most important one of all because the two most important Uniform Acceleration formulae can be deduced from it.When a velocity vs time graph of an object's motion looks like a horizontal line the object's acceleration is zero, Vf = Vi and usually the most basic formulae for motion are used.