need help on a particular question of my assignment...

Using strobe photography a badminton shuttle's path is followed. the strobe light takes photos every 0.2 seconds and lasts for a total of 3.2 seconds.

The actual positions of the shuttle are below...

Horizontal - 0.00 0.80 1.56 2.29 2.97 3.61 4.21 4.76 5.28 5.75 6.18 6.59 6.97 7.33 7.67 8.00 8.32

Vertical - 0.00 2.24 4.08 5.56 6.71 7.54 8.09 8.36 8.38 8.16 7.68 6.97 6.05 4.93 3.64 2.19 0.59

i plotted a graph of the horizontal position against time.

i am then asked to find the horizontal velocity of the shuttle using the gradient of the previous graph. it is supposebly a linear function and i just need to differentiate it. i think it may be y=ax^2+bx+c...i need a table and a graph.

not completely sure tho. help would be great. thanks.

2. If we're talking about ONLY the horizontal position against time, we do get a function that is nicely approximated by a quadratic.

I get x(t) = -0.4373t^2 + 3.9568t + 0.0490
The derivative of this is x'(t) = -0.8746t + 3.9568
This gives the first three pieces of velocity
x'(0) = 3.9568
x'(1) = 3.0822
x'(2) = 2.2076

On the other hand, the first three first differences of the given data suggest:

(0.8-0)/0.2 = 4
(1.56-0.8)/0.2 = 3.8
(2.29-1.56)/0.2 = 3.65

There is some discrepancy there. If you can come to terms with exactly when the two versions are measured, perhaps you can reconcile them.

Hint: Derivative are instantaneous and differences are averages.

Thanks!