The diagram is a distance time graph for a train journey from A to B .
Let's say I want to find the speed of the train at 08 08 .
How do I go about doing this ?
Why I can't take the total distance travelled by the train from 0800-0808 the divide it by 8 mins ?
What you suggest would calculate the average speed over that time.
To calculate the instantaneous speed, find the slope of the tangent to the curve at the time you want.
One way to find the "instantaneous" speed is this: take a small pocket or purse mirror and hold it across the graph at the given time. Look at the graph and its reflection in the mirror. You will see a sharp "angle" where the graph meets its image. Slowly rotate the mirror until graph appears to go smoothly into its image. Hold the mirror in place while you use the it to draw a line. That line will be perpendicular to the curve. Now do the same thing holding the mirror over this perpendicular that you just drew and rotate the mirror until that line appears to merge smoothly into its image. Again, hold the mirror draw a line along the mirror. This line will be tangent to the curve. Find the slope of that line.
You can approach such curve as a composition of two quadratic functions. The first represents the accelerating path and the other the decelerating path.
$$\text{y}_{1} = {a}_{1} \cdot x^2$$
$$\text{y}_{2} = {a}_{2} \cdot x^2 + {b}_{2} \cdot x + {c}_{2} $$
Subject to:
$$\text{y'}_{1} = 0 @ 8:00$$
$$\text{y}_{1} = \text{y}_{2} @ 8:13 aprox$$
$$\text{y'}_{1} = \text{y'}_{2} @ 8:13 aprox$$
$$\text{y'}_{2} = 0 @ 8:27:30$$
${a}_{1}$ must be positive and ${a}_{2}$ must be negative.
Then derivate two times.
Remember to use real numbers for time.
8:08 = 8+8/60 = 8.0133...
Your question:
Why I can't take the total distance travelled by the train from 0800-0808 the divide it by 8 mins ?
Doing this, you are averaging speed.
