AVERAGE RATE OF CHANGE AND INSTANTANEOUS RATE OF CHANGE

Problem 1 :

A point moves along a straight line in such a way that after t seconds its distance from the origin is

s  =  2t² + 3t meters.

(i) Find the average velocity of the points between t = 3 and t = 6 seconds.

(ii) Find the instantaneous velocities at t = 3 and t = 6 seconds.

Solution :

(i)  Average rate of change between t = 3 and t = 6

=  [f(6) - f(3)] / (6-3)  ---(1)

f(x)  =  2t² + 3t meters.

f(6)  =  2(6)²+3(6)  ==>  72+18  ==>  90

f(3)  =  2(3)²+3(3)  ==>  18+9  ==>  27

Applying in (1), we get

=  (90-27)/(6-3)

=  63/3

=  21 meter/.second

(ii)  f(x)  =  2t² + 3t

f'(x)  =  4t+3

Problem 2 :

A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of

s  =  16t² in t seconds.

(i) How long does the camera fall before it hits the ground?

(ii) What is the average velocity with which the camera falls during the last 2 seconds?

(iii) What is the instantaneous velocity of the camera when it hits the ground?

Solution :

(i)  s  =  16t²

To reach the ground it has to cover 400 meter.

16t²  =  400

t²  =  400/16  =  25

t  =  5 sec

So, it takes 5 seconds to reach the ground.

(ii) average velocity with which the camera falls during the last 2 seconds.

Last 2 seconds means  t  =  3 to t  =  5

f(t)  =  16t²

f(3)  =  16(3)²  ==>  144

f(5)  =  16(5)²  ==>  400

Average velocity  =  [f(5) - f(3)] / (5-3)

=  (400-144)/2

=  256/2

=  128 ft/sec

(iii)  Instantaneous rate of change 

=  Instantaneous velocity

f(t)  =  16t²

f'(t)  =  32t

f'(5)  =  32(5)

f'(5)  =  160 ft/sec

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