AVERAGE RATE OF CHANGE WORD PROBLEMS

Problem 1 :

The height of a model rocket in flight can be modelled by the equation

h(t) = -4.9t² + 25t + 2

where h is the height in meters at t seconds. Determine the average rate of change in the model rocket’s height with respect to time during 

a. the first second

b. the second second

Solution :

a) Calculate the height at t = 0 seconds and t = 1 seconds.

Average rate of change = [h(1) - h(0)] / (1 - 0)

Average rate of change at the first second = (22.1 - 2)/1

= 20.1 meter/second

b) 

Calculate the height at t = 1 seconds and t = 2 seconds.

Average rate of change = [h(2) - h(1)] / (2 - 1)

Average rate of change at the second second = (32.4 - 22.1)/(2 - 1)

= 10.3 meter/second

Problem 2 :

A pebble is dropped from a cliff, 80 m high. After t seconds, the pebble is s meters above the ground, where s(t) = 80 - 5t², 0 ≤ t ≤ 4

a. Calculate the average velocity of the pebble between the times t = 1 s and t = 3 s.

b. Calculate the average velocity of the pebble between the times t = 1 s and t = 1.5 s.

Solution :

s(t) = 80 - 5t²

a) 

= [s(3) - s(1)] / (3 - 1)

= (35 - 75)/2

= -40/2

The average velocity in this 2 s interval is = -20 meter/seconds

b) 

= [s(1.5) - s(1)] / (1.5 - 1)

= (68.75 - 75)/0.5

= -6.25/0.5

The average velocity in this 0.5 s interval is = - 12.5 meter/seconds

Problem 3 :

A tank holds 500 L of liquid, which takes 90 min to drain from a hole in the bottom of the tank. The volume, V, remaining in the tank after t minutes is

V(t) = 500(1 - t/90)² where 0 ≤ t ≤ 90

a. How much liquid remains in the tank at 1 h?

b. What is the average rate of change of volume with respect to time from 0 min to 60 min?

c. How fast is the liquid draining at 30 min?

Solution :

V(t) = 500(1 - t/90)² where 0 ≤ t ≤ 90

a) 

t - number of minutes

1 hour = 60 minutes

V(60) = 500(1 - 60/90)²

= 500(1-0.66)²

= 500 (0.33)²

= 54.5

Approximately 55 Liters

b) 

V(0) = 500(1 - 0/90)²

= 500(1)²

= 500 Liters

V(60) = 57.8 liters

Average rate of change from 0 minutes to 60 minutes = [V(60) - V(0)]/(60 - 0)

= (57.8 - 500)/60

= 7.37

Approximately 7.4 liters.

c) By finding the rate of change at t = 30, we will get how fast it is decreasing.

V'(t) = 1000(1 - t/90)(1/90)

= (100/9)(1 - t/90)

When t = 30, we get

= (100/9)(1 - 30t/90)

= (100/9)(2/3)

= 2000/27 liters/minute

Problem 4 :

Volume of the sphere is given by

V(r) = (4/3)πr³

a. Determine the average rate of change of volume with respect to radius as the radius changes from 10 cm to 15 cm.

b. Determine the rate of change of volume when the radius is 8 cm.

Solution :

V(r) = (4/3)πr³

a)

Average rate of change = (13500π/3 - 4000π/3)/(15 - 10)

= (13500 - 4000)π/15

= 9500π/15

= 1900π/3 cm³/cm

b) Since we find the rate of change at the specific point, we find instantaneous rate of change.

V'(r) = (4/3)(3πr²)

= 4πr²

When r = 8

= 4π(8)²

= 256π cm³/cm

Problem 5 :

The population, P, of a bacteria colony at t hours can be modelled by

P(t) = 100 + 120t + 10t² + 2t³

a. What is the initial population of the bacteria colony?

b. What is the population of the colony at 5 h?

c. What is the growth rate of the colony at 5 h?

Solution :

P(t) = 100 + 120t + 10t² + 2t³

a) Initial population is at t = 0

P(0) = 100 + 120(0) + 10(0)² + 2(0)³

= 100 bacteria

b) Population at 5 hours

P(5) = 100 + 120(5) + 10(5)² + 2(5)³

= 100 + 600 + 250 + 250

= 1200 bacteria

c) Finding the instantaneous rate of change, we find the growth rate at 5 hours.

P'(t) = 0 + 120(1) + 20t + 6t²

= 120 + 20t + 6t²

Applying t = 5, we get

= 120 + 20(5) + 6(5)²

= 120 + 100 + 150

= 370 bacteria