You ran approximately 5278 feet. This question only involving converting units. 1 mile=1.609km=1609m. 1m=3.28 feet. So 1 mile=1609*3.28=5278 feet approximately.
Answer:
The absolute maximum is
and the absolute minimum value is 
Step-by-step explanation:
Differentiate of
both sides w.r.t.
,


Now take 



![\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]](https://tex.z-dn.net/?f=%5CRightarrow%201-2%5Csin%20%5E2t%20%3D%5Csin%20t%20%20%5Cquad%20%5Cquad%20%20%5B%5Cbecause%20%5Ccos%202t%20%3D%201-2%5Csin%20%5E2t%5D)






In the interval
, the answer to this problem is 
Now find the second derivative of
w.r.t.
,

![\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cleft%5Bf%27%27%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%20%7B%5Cpi%7D6%7D%3D-2%5Ctimes%20%5Cfrac%20%7B%5Csqrt%203%7D2-4%5Ctimes%20%5Cfrac%7B%5Csqrt%203%7D2%3D-3%5Csqrt%203)
Thus,
is maximum at
and minimum at 
![\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0](https://tex.z-dn.net/?f=%5Cleft%5Bf%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%20%7B%5Cpi%7D6%7D%3D2%5Ctimes%20%5Cfrac%20%7B%5Csqrt%203%7D2%2B%5Cfrac%7B%5Csqrt%203%7D2%3D%5Cfrac%7B3%5Csqrt%203%7D2%5C%3B%5Ctext%7Band%7D%5C%3B%5Cleft%5Bf%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%7B%5Cpi%7D2%7D%3D%202%5Ctimes%200%2B0%3D0)
Hence, the absolute maximum is
and the absolute minimum value is
.