Emily is observing the velocity of a cyclist at different times. After four hours, the velocity of the cyclist is 14 km/h. After nine hours, the velocity of the cyclist is 4 km/h.

Part A: Write an equation in two variables in the standard form that can be used to describe the velocity of the cyclist at different times. Show your work and define the variables used. (5 points)

Part B: How can you graph the equations obtained in Part A for the first 10 hours? (5 points)

Assume that Emily starts from rest so that the velocity is zero at 0 hours.

Define the function v(t) for velocity in km/h, and t in hours.

Part A
From t=0 to t=4, v changes from 0 to 14.
Let v(t) = mt + b
The slope is
m = 14/4 = 3.5, and the y-intercept is
b = 0.
Therefore
v(t) = 3.5t, for 0 ≤ t ≤ 4

Fromt=4 to t=9, changes from 14 to 4.
Let v(t) = mt + b
The slope is
m = (4 - 14)/(9 - 4) = -2
Therefore
v = -2t + b
When t=4, v=14, therefore
14 = -2*4 + b = -8 + b
22 = b
v(t) = -2t + 22, for 4 < t ≤ 9.

Answer:
The equation for the velocity is
v(t) = 3.5t, 0 ≤ t ≤ 4
= -2t + 22, 4 < t ≤ 9

Part B
The equation for velocity is a piecewise function that is graphed as shown below. If we assume that the second part of the equation is valid at t=10, then
v(10) = -2*10 + 22 = 2 km/h