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An athlete completes one round of a circular track of diameter $200\, m$ in $40\, s$. What will be the distance covered and the displacement at the end of $2 \,min$ $20 \,s$ ?
$Distance \;covered=2200 \,m\;;\;Displacement=200 \,m$
$Distance \;covered=\frac{4400}{7} \,m\;;\;Displacement=200 \,m$
$Distance \;covered=\frac{4400}{7} \,m\;;\;Displacement=100 \,m$
$Distance \;covered=2200 \,m\;;\;Displacement=100 \,m$
Solution
Diameter of circular track $( D )=200\, m$
Radius of circular track $(r)=200 / 2=100 \,m$
Time taken by the athlete for one round $(t)=40\,s$
Distance covered by athlete in one round $(s)=2 \pi r$
$=2 \times(22 / 7) \times 100$
Speed of the athlete $(v)=$ Distance $/$ Time
$=(2 \times 2200) /(7 \times 40)$
$=4400 / 7 \times 40$
Therefore, Distance covered in $140 \,s =$ Speed $( s ) \times$ Time $(t)$
$=4400 /(7 \times 40) \times(2 \times 60+20)$
$=4400 /(7 \times 40) \times 140$
$=4400 \times 140 / 7 \times 40$
$=2200 \,m$
Number of round in $40 \,s =1$ round
Number of round in $140 \,s =140 / 40 =3^{1 / 2}$
After taking start from position $X$,the athlete will be at postion $Y$ after $3^{1 / 2}$ rounds as shown in figure.
Hence, Displacement of the athlete with respect to initial position at $X= XY$
$=$ Diameter of circular track
$= 200\, m$
