A ball thrown in the air has a height of y = – 16x² + 50x + 3 feet after x seconds. a) What are the units of measurement for the rate of change of y? b) Find the rate of change of y between x = 0 and x = 2?
A ball thrown in the air has a height of y = – 16x² + 50x + 3 feet after x seconds. a) What are the units of measurement for the rate of change of y? b) Find the rate of change of y between x = 0 and x = 2?
Answer:
(a) ft/s
(b) 1ft/s
Step-by-step explanation:
Given equation;
y = (- 16x² + 50x + 3)ft ————-(i)
Where;
y is measured in feet(ft)
x is measured in seconds(s).
(a) The rate of change of y with respect to x is found by dividing the total change in y by the total change in x. i.e
Δy / Δx
Where;
Δy = y₂ – y₁
Δx = x₂ – x₁
∴ Δy / Δx = [tex]\frac{y_2 – y_1}{x_2 – x_1}[/tex] ————–(ii)
Since y is measured in feet, Δy will also be measured in feet.
Also, since x is measured in seconds, Δx will also be measured in seconds.
Therefore, the rate of change of y with respect to x (Δy / Δx) will be measured in feet per second (ft/s)
(b) The rate of change of y between x = 0 and x = 2 can be found by using equation (ii)
Where;
y₂ is the value of y at x = 2 found by substituting x = 2 into equation (i)
y₁ is the value of y at x = 0 found by substituting x = 0 into equation (i)
=> y₂ = – 16(2)² + 50(2) + 3 = 39
=> y₁ = – 16(1)² + 50(1) + 3 = 37
Now, substitute the values of y₂, y₁, x₂ and x₁ into equation (ii)
Δy / Δx = [tex]\frac{39 – 37}{2 – 0}[/tex]
Δy / Δx = [tex]\frac{2}{2}[/tex]
Δy / Δx = 1
Therefore, the rate of change of y is 1 ft/s