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Paul’s account balance is less than -$50. What is a possible balance for paul’s account? Explain.
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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
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