Question

7. Along the eastern shore of Lake Michigan from Lake Macatawa (near Holland) to Grand Haven, there is a bike bath that runs almost directly north-south. For the purposes of this problem, assume the road is completely straight, and that the function s(t) tracks the posi- tion of the biker along this path in miles north of Pigeon Lake, which lies roughly halfway between the ends of the bike path. Suppose that the biker's velocity function is given by the graph in Figure 4.1.11 on the time a. Approximately how far north of Pigeon Lake was the cyclist when she was the greatest b. What is the cyclist's total change in position on the time interval 0 s t s 2? At 2, c. What is the total distance the biker traveled on0st 4? At the end of the ride, how interval 0 t s 4 (where t is measured in hours), and that s(0) 1 distance away from Pigeon Lake? At what time did this occur? was she north or south of Pigeon Lake? close was she to the point at which she started? 218 4.1 Determining distance traveled from velo 10 y 10 hrs hrs -6 -6 10 10 Figure 4.1.11: The graph of the biker's velocity, y(t), at left. At right, axes to plot approximate sketch of y s(t). d. Sketch an approximate graph of ys(t), the position function of the cyclist, on t interval 0 St 4. Label at least four important points on the graph of s

Answer 1

$v(t)=11\,mi/h$ for $0<t<1$

$v(t)=-10\,mi/h$ for $1<t<3$

$v(t)=9\,mi/h$ for  $3<t<4$

a) For time interval $0<t<1$   $s(t)=1&plus;11t$ ,when $t=1$, position $s=12\,mi$

For time interval $1<t<3$   $s(t)=12-10(t-1)$ when $t=3$ ,position is $s=-8\,mi$

For time interval $3<t<4$     $s(t)=-8&plus;9(t-3)$ when $t=4$ , position is $s=1\,mi$

Greatest distance from starting point ( pigeon lake ) is $s=12\,mi$ at time $t=1\,hr$

b) At time $t=0\,hr$ , she is at $s(0)=1\,mi$

At time $t=2\,hr$ , she is at $s(2)=12-10(2-1)=2\,mi$

Total change in position of biker in $0\leq t\leq 2$ is $\Delta s=s(2)-s(0)=2-1=1\,mi$

She is north of pigeon lake.

c) Distance traveled in first 1 hour ($0<t<1$) is $|s(1)-s(0)|=11\,mi$

Distance traveled in next 2 hours ($1<t<3$ ) is $|s(3)-s(1)|=|-8-12|=20\,mi$

Distance traveled in next one hour ($3<t<4$) is $|s(4)-s(3)|=|1-(-8)|=9\,mi$

Total distance traveled is $s_{total}=11&plus;20&plus;9=40\,mi$

To know her final position , $s(4)=-8&plus;9(4-3)=1\,mi$

That is she is right where she started.

d)

2 mph y )10 4 hrs hrs 2 4 5 21123 /45 3 -10

point 1( in green color) is her starting point, point 2 is her position at the end of one hour, point 3 is her position at the end of 3 hours and point 4 is her position at the end of 4 hours.