RELATED RATES WORD PROBLEMS INVOLVING AVERAGE VELOCITY

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 velocity at t = 3 and t = 6 seconds.

Solution :

Average velocity = [f(b) - f(a)]/(b - a)

Substitute a = 3 and b = 6.

Average velocity = (90 - 27)/(6 - 3)

= 63/3

= 21

Hence the average velocity is 21 meter/sec.

(ii) The instantaneous velocity is given by s'(t) = 4t + 3.

Instantaneous velocity at t = 3 sec is

s'(3) = 4(3) + 3

= 15 meter/sec

Instantaneous velocity at t = 6 sec is

s'(6) = 4(6) + 3

= 27 meter/sec

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) Given s = 16t² ----(1).

The camera has to fall 400 ft to hit the ground.

To calculate the time taken by the camera before it hits the ground, substitute s = 400 in (1).

400 = 16t²

t² = 25, then t = 5 sec.

(ii) Since the total time is 5 seconds, the average velocity with which the camera falls during the last 2 seconds is given by

S(5) = 16(5)² = 400

S(3) = 16(3)² = 144

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

= (400 - 144) / 2

= 256/2

The average velocity is 128.

(iii) When the camera hits the ground time t = 5. The instantaneous velocity is given by

s'(t) = 32t

= 32(5)

= 160

The instantaneous velocity is 160 feet/sec.

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