1. Activity 5-2: Understanding Rates of Change

2. Activity 5-2: Understanding Rates of Change
You stand on a cliff that is 30 metres above Lake Superior. You decide to drop a rock to see how long it takes to hit the water. Due to gravity, the rock height in relation to time follows the function:
h(t) = 30 – 4.9t2
What graph best describes this model?
LINEAR
QUADRATIC
EXPONENTIAL
SINUSOIDAL

3. Activity 5-2: Understanding Rates of Change
How long did it take the rock to hit the water to 2 decimal places?
The graph of the function is given below:
Solution:
We can factor h(t) = 30 – 4.9t2
h(t) = -4.9(t2 – 6.122)
h(t)=0 when the rock hits water
0= -4.9(t2 – 6.122)
.: t2=6.122 or t = (only positive)
h(t) = 30 – 4.9t2
What is the average speed of the rock from the time it was dropped to the time it hits the water?
When the rock hits the water at 2.47, the height of the rock is 0 m
Solution:
Speed is the distance travelled in 2.47 seconds
s = d/t = 30m/2.47s
s = m/s
2.47

4. Activity 5-2: Understanding Rates of Change
The graph of the function is given below:
The average speed of the rock is represented by the slope of the line going through the initial point and the final point
(0, 30)
h(t) = 30 – 4.9t2
VIEW SLOPE
The line joining the two points is called a SECANT. A secant is a line that passes through two points
SECANT
The average speed is represented by the slope of the secant.
SLOPE OF SECANT
(2.47, 0)

5. Activity 5-2: Understanding Rates of Change
The graph of the function is given below:
h(t) = 30 – 4.9t2
Find the average speed of the rock between 1 and 2.47 seconds.
VIEW ANSWER
FIND HEIGHT OF ROCK AT t=1 and t=2.47
We need to determine the height of the rock at 1 and 2.47 seconds using the equation:
h(1) = 30 – 4.9(1)2
h(1) = 25.1 m
h(2.47) = 0 m
(1, 25.1)
SECANT
FIND THE SPEED USING THE SLOPE OF THE SECANT
We can draw the secant between the two points. Using the slope equation we get:
(2.47, 0)

6. Activity 5-2: Understanding Rates of Change
The graph of the function is given below:
h(t) = 30 – 4.9t2
Find the average speed of the rock between 2 and 2.47 seconds.
VIEW ANSWER
FIND HEIGHT OF ROCK AT t=2 and t=2.47
We need to determine the height of the rock at 2 and 2.47 seconds using the equation:
h(2) = 30 – 4.9(2)2
h(2) = 10.4 m
h(2.47) = 0 m
SECANT
(2, 10.4)
FIND THE SPEED USING THE SLOPE OF THE SECANT
We can draw the secant between the two points. Using the slope equation we get:
(2.47, 0)

7. Activity 5-2: Understanding Rates of Change
Part 1: Review
The graph of the function is given below:
h(t) = 30 – 4.9t2
Let us view the average speed of the rock from the various times to when it hits ground.
VIEW SECANTS
-22.1 m/s
-17.07m/s
-12.15m/s
How would you go about estimating the rock’s speed when it hit the water?
SHOW ANSWER
Bring the point closer and closer to t=2.47 seconds. Find the slope of the secant.
(2.47, 0)

8. Activity 5-2: Understanding Rates of Change
Part 1: Review
The graph of the function is given below:
h(t) = 30 – 4.9t2
Let us view how this occurs
VIEW SECANTS
How can we find the velocity of the rock the instant it hits the water?
VIEW ANSWER
Slope of Secant:
-25.23
Slope of Secant:
-24
Slope of Secant:
-23.26
We must find the instantaneous velocity which represents the slope of a point. The slope of a point is called a TANGENT
(2.2, 6.28)
(2.3, 4.08)
(2.4, 1.78)
(2.47, 0)