High school pre cal sandra can run a mile in 6 minutes. at her high school track, shown below, 4 laps around the track equals 1 mile. if sandra starts at point s, which parametric equations will find her position on the track at any time t in seconds?

We know that
1) Sandra can run a mile in 6 minutes> 6*60> 360 sec
2) 4 laps around the track equals 1 mile
so
4 laps around the track in 360 sec
1 lap in 360/4> 90 sec
3) the position of Sandra for t=90 sec must be equal to the point S (0,56)
I proceed to analyze each case for t=90 sec

case a) x(t)=-140 cos(pi*t/45)    y(t)=112 sin(pi*t/45)
x(t)=-140 cos(pi*90/45)> -140
y(t)=112 sin(pi*90/45)> 0
the position is the point (-140,0)> is not the point S

case b) x(t)=140 sin(pi*t/90)  y(t)=-112 cos(pi*t/90)
x(t)=140 sin(pi*90/90)> 0
y(t)=-112 cos(pi*90/90)> 112
the position is the point (0,112)> is not the point S

case c) x(t)=-70 sin(pi*t/45)  y(t)=56 cos(pi*t/45)
x(t)=-70 sin(pi*90/45)> 0
y(t)=56 cos(pi*90/45) > 56
the position is the point (0,56)> is equal to the point S> is the solution

case d) x(t)=70 cos(pi*t/90)  y(t)=-56 sin(pi*t/90)
x(t)=70 cos(pi*90/90)> -70
y(t)=-56 sin(pi*90/90)> 0
the position is the point (-70,0)> is not the point S

 therefore

the answer is the option C
x(t)=-70 sin(pi*t/45)  y(t)=56 cos(pi*t/45)

The answer that i got was 8/3