Given:
A cyclist pedals at a rate of 300 m min(exponent of -1) for 20 minutes. Then she slows down to 150 m min (exponent of -1) for 16 minutes, then races at 400 m min (exponent -1) for four minutes.
We will draw the diagram between the rate (speed of the cyclist) and the time (t) in minutes, the graph will be as follows:
For the first 20 minutes, the rate increased from 0 to 300 m/min
Then the next 16 minutes, the rate decreased to 150 m/min
And in the last 4 minutes, the rate increased to 400 m/min.
To find the distance traveled, we will find the area under the lines according to a specific time.
So, first, we will find the distance traveled after 20 minutes
It will be as follows = d(20)
[tex]d(20)=\frac{1}{2}*300*20=3000\text{ }m=3\text{ }km[/tex]And the distance traveled from 20 min to 36 min is the area of a trapezoid with a height = 16 min. and the parallel base are 150 and 300
So, the area will be =
[tex]\frac{1}{2}(300+150)*16=3600\text{ }m=3.6\text{ }km[/tex]So, the distance traveled after 36 minutes = 3 + 3.6 = 6.6 km
And the distance traveled from 36 min to 40 min is the area of a trapezoid with a height = 4 min. and the parallel bases are 150 and 400
So, the area =
[tex]\frac{1}{2}(150+400)*4=1100\text{ }m=1.1\text{ }km[/tex]So, the total distance after 40 minutes = 6.6 + 1.1 = 7.7 km.
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To find the piecewise linear function for the distance of D(t) in terms of time (T) in minutes.
First, we will write the function v(t) that represents the rate from the graph.
[tex]v(t)=\begin{cases}{15t\rightarrow0\leq t\leq20} \\ -9.375t+487.5\rightarrow20\leq t\leq36 \\ {62.5t-2100\rightarrow t\ge36}\end{cases}[/tex]To find the function of the distance integrate each function with respect to the time t:
[tex]D(t)=\begin{cases}{7.5t^2}\rightarrow0\leq t\leq20 \\ {-4.6875t^2+487.5t-4875\rightarrow20\leq t\leq36} \\ {31.25t^2-2100t+41700\rightarrow t\ge36}\end{cases}[/tex]
