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Intuitive Idea of Derivatives

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Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of 4.9`t^2` metres in t seconds, i.e., distance s in metres covered by the body as a function of time t in seconds is given by 
s = 4.9`t^2`. 
In the following table the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff.
The objective is to find the veloctiy of the body at time t = 2 seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at t = 2 seconds and hope that these throw some light on the velocity at t = 2 seconds. 
Average velocity between t = `t_1` and t = `t_2` equals distance travelled between t = `t_1` and t = `t_2`  seconds divided by  `(t_2 – t_1)`. Hence the average velocity in the first two seconds.

t s
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

= `("Distance travelled between"  t_2 = 2 and t_1 = 0)/("Time interval" (t_2-t_1))`
= `((19.6 - 0)m)/((2-0)s)` = 9.8 m/s
Similarly, the average velocity between t = 1 and t = 2 is  
`((19.6 - 4.9)m)/((2-1)s)` = 14.7 m/s
we compute the average velocitiy between t = `t_1` and t = 2 for various `t_1`. gives the average velocity (v), t = `t_1` seconds and t = 2 seconds

`t_1` 0 1 1.5 1.8 1.9 1.95 1.99
v 9.8 14.7 17.15 18.62 19.11 19.355 19.551

From the above table, The average velocity is gradually increasing.  the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2.Hoping that nothing really dramatic happens between 1.99 seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just above 19.551 m/s. 

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