NZ Level 7 (NZC) Level 2 (NCEA) Relating gradient and original functions (investigation) - including velocity time to position time
Lesson

## Introduction

Consider an object moving along a path. Maybe it is a car on a road network or an insect looking for food or a comet in the solar system. We might track the position of the object along its path at regular small time intervals.

If the distance moved along the path is seen to vary between one time interval and the next, we say the speed of the object is changing. The object has a speed related to the distance moved along the path in a single time interval. That is, speed is understood to be the time rate of change of position along the path.

A related concept is that of velocity.  This is the rate of change of position along a line. It takes into account both the distance travelled in a small time interval and also the direction of motion.

In tracking the position of an object moving along a path, we might create a table relating times to distances travelled. This information could then be displayed graphically using the Cartesian plane with times on the horizontal axis and the corresponding distances travelled on the vertical axis.

#### An example

 time (days) distance (,000 Km) $0$0 $1$1 $2$2 $3$3 $4$4 $5$5 $6$6 $7$7 $8$8 $1$1 $3$3 $4$4 $6$6 $9$9 $11$11 $12$12 $12.5$12.5 $13$13

We have plotted the distance-time points on the graph below. The points from the table have been connected with lines on the assumption that the distances vary continuously between the recorded times. (There are no sudden jumps in distance.)

It is clear from the table and from the graph that the distances travelled over the equal time intervals are not the same. Thus, the slopes of the connecting lines vary and the varying slopes correspond to changes in the speed of the object. We can calculate average speeds over each day. For the first day, the distance travelled was $3000$3000 km. this is $\frac{3000}{24}=125$300024=125 kilometres per hour. In day six the average speed was $\frac{12000-11000}{24}\approx41.67$12000110002441.67 km/h.

Because the distances were measured only once per day, the graph appears to have sudden changes of slope. In reality, the speed of the object could vary smoothly and we would obtain a more accurate graphical representation of the motion by measuring the distance over shorter time intervals, say half a day or one hour.

In the limit, if the time intervals could be made very short, the graph would be perfectly smooth and we could speak of the instantaneous speed of the object as the gradient of the smooth curve at the point of interest.

The speedometer of a car measures these instantaneous speeds.

As was done in the case of the table of distance-time data, we could record a sequence of instantaneous speeds of a motor vehicle or some other object at regular time intervals and then make a graph showing how the speed varies with time.

The time rate of change of speed is the acceleration. It corresponds to the gradient of the speed-time graph.

In physics, an acceleration is understood to be the rate of change in the velocity of an object along a line. As with velocity, acceleration has both magnitude and direction. However, in everyday usage acceleration is understood more loosely to be any change in speed, without reference to the object's direction.

### challenge 1

The following graph shows the same points as in the graph above but without the connecting lines. • Make a more realistic distance-time graph by carefully drawing a smooth curve connecting the plotted points.
• Now construct tangent lines at days $1$1 to $7$7. The slopes of these tangent lines represent the instantaneous speeds at each of these times. By careful measurement of the gradients, determine the speeds.
• Make a table relating times and speeds and use this data to construct a speed-time graph by drawing a smooth curve connecting the plotted points.
• Investigate how the acceleration of an object varies by considering the changes in the gradient of the speed-time graph.

If you have completed Challenge $1$1 above, you will have seen that the speed-time graph comes from plotting the smoothly varying gradients of a distance-time graph. It is a powerful idea that it is also possible to reverse the process and deduce the shape of a distance-time graph from a speed-time graph.

These ideas form the basis of the differential and integral calculus formulated by Isaac Newton and Gottfried Leibniz in the seventeenth century.

To see how the 'reverse' process works, consider a small slice of a speed-time graph, corresponding to a brief time interval of length $\Delta t$Δt. (The Greek letter $\Delta$Δ indicates a small increment in the variable $t$t.)

During the time interval $\Delta t$Δt, the object moves a distance $\Delta s$Δs. We have seen above that the average speed $v$v of the object is the slope of the yet to be drawn distance-time graph over the time $\Delta t$Δt. That is, $v=\frac{\Delta s}{\Delta t}$v=ΔsΔt and therefore, $\Delta s=v\Delta t$Δs=vΔt where $v$v is the average speed during the small interval $\Delta t$Δt. (Note that in the speed-time graph, the average speed $v$v over an interval has to be somewhere between the speed at the beginning of the interval and the speed at its end.)

But, the quantity $\Delta s=v\Delta t$Δs=vΔt is just the area of the thin strip below the speed-time graph. It follows that the distance travelled at any elapsed time $t$t is just the cumulative total up to that point of the areas of all the thin strips that could be drawn beneath the graph. ### challenge 2

• Make a table showing speed versus time, much like the table of distance versus time shown above. Choose whatever units you think would be appropriate and make up some non-negative speed values to complete your table.
• Plot the points from the table to make a speed-time graph.
• Choose a starting time $t_0$t0 and an end-time. The goal will be to obtain a good estimate of the distance travelled by the object over that time interval.
• Divide your time interval into a few reasonably thin strips like the strip shown in the example above. For each strip, estimate the average speed over that particular $\Delta t$Δt.
• Find the areas of each of the strips by multiplying the average speed for each interval by its duration. Sum the areas to obtain an estimate of the total distance travelled by the object.
• You could sum the areas in a step-by-step way to obtain the distances travelled at several points in time. It should then be possible to draw a distance-time graph corresponding to your original speed-time graph and to compare the shapes of the two graphs.

### Outcomes

#### M7-9

Sketch the graphs of functions and their gradient functions and describe the relationship between these graphs

#### 91262

Apply calculus methods in solving problems