Why are avg. v and instant. v different?

For instantaneous velocity we take the limit as Δt approaches to zero.
For average velocity we can take Δt for any duration.

 

So i could not understand what that b represents....

But simply, if u have displacement as a function of time, u can use

average speed = total displacement/total time

and instantaneous velocity = df(t)/dt
that is differentiate ur (displacement as a function of time) function to get instantaneous velocity

 

The speed (obviously) changes at certain times, due to acceleration. The average velocity takes the average speed at all points divided by the time (makes sense right?). The instantaneous velocity takes the exact point, therefore it's the derivative of the displacement.

 

This is similar to mean value theorem. For any continous function x(t) over the interval {a, b}, there is at least one point c within the interval {a,b} where x'(c) = (x(b) - x(a)) / (b - a). This means there is some point c where the instantaneous velocity x'(c) equals the average velocity.

For the second equation

x(t) = 2 t³ - 15 t + 20

the average velocity over {0,t} is

(x(t) - x(0)) / (t - 0) = 2 t² - 15

define a so that (a t) = c, the value we're looking for

the instantaneous velocity is

x'(a t) = 6 (a t)² - 15

then find a so that instantaneous velocity = average velocity
x'(a t) = (x(t) - x(0)) / (t - 0)

6 (a t)² - 15 = 2 t² - 15
6 (a)² t² = 2 t²
6 (a)² = 2
(a)² = 1/3
a = sqrt(1/3)

so c = sqrt(1/3) t

This means that for x(t) = 2 t³ - 15 t + 20, the instantaneous velocity at sqrt(1/3) t equals the average velocity over {0,t}.

You've already noted that for x(t) = 2 t³ - 15 t + 20, the instantaneous velocity at (1/2) t equals the average velocity over {0,t}.