Question

An athelete
athelete runs in a circular track
of diameter 100m completes one round
in 20s. calculate
distance covered at the end of 1 min
20s
​

Answer 1

Question

An athelete runs in a circular track

of diameter 100m completes one round

in 20s. calculate distance covered at the end of 1 min 20s.

Solution-

An athlete completes one round of a circular track of diameter 100m in 20s.

Given that, the diameter is 100 m. So, the radius is 50 m (half of diameter).

As the athlete covers the path in a circular motion. So, the displacement is zero.

Distance cover in one round (during 20 seconds) = 2πr

(2πr = Circumference of the circle)

→ 2 × 22/7 × 50

→ 2200/7

→ 314.28 m

Distance covered in 1 second (during first or one round) = 314.28/20 = 15.71 m

We have to find the distance covered at the end of 1min 20s.

Distance covered in 1min 20sec or 80 sec = 15.71 × 80 = 1256.8 m

Answer 2

${ \sf{ { \huge{ \red{ \underline{ \underline{Question:-}}}}}}}$



▪ An athlete runs in a circular path of diameter 100m completes one round in 20 sec . Calculate the distance covered at the end of 1min 20 sec ??



${ \red{ \sf{ \huge{ \underline{ \underline{Solution:-}}}}}}$



${ \bold{ \underline{ \purple{GIVEN- }}}}$



${ \sf{time \: taken \: for \: completing \: 1 \: round = 20 \: sec}}$



${ \sf{ {diameter \: of \: the \: circular \: track \: = 100 \: m}}}$



${ \sf{ \rightarrow{ \boxed{ \green{ radius = \frac{diameter}{2}}}}}}$



therefore,



${ \red{ \sf{radius \: of \: the \: circular \: track = \frac{100 \: m}{2}}}}$



${ \implies{ \sf{radius = 50 \: m}}}$



${ \bold{ \underline{for \: 1 \: round}}}$



▪ Distance covered by the athlete



= circumference of the circular track



${ \tt{ \red{distance = 2 \: \pi \: radius}}}$




${ \implies{ \sf{ \blue{distance = 2 \: \pi \: \times 50m}}}}$



${ \implies{ \sf{ \blue{distance = 100 \: \pi \: m}}}}$



${ \red{ \tt{time = 20 \: sec}}}$



${ \boxed{ \pink{ \boxed{ \rm{speed = \frac{distance}{time}}}}}}$



${ \rightarrow{ \sf{ \blue{speed = \frac{100 \: \pi \: m}{20 \: sec}}}}}$



${ \implies{ \sf{ \blue{ \underline{speed = 5 \: \pi \: m \: {sec}^{ - 1}}}}} }$



${ \bold{ \underline{ \purple{to \: find}}}}$



▪ Distance covered by the athlete at the end of 1 min 20 sec???



${ \red{ \tt{time = 1 \: min \: 20 \: sec}}}$



▪ [ 1 minute = 60 seconds ]



${ \sf{ \red{time = 60 \: sec + 20 \: sec = 80 \: sec}}}$



${ \tt{ \red{speed = 5 \: \pi \: m \: {sec}^{ - 1}}}}$



${ \boxed{ \boxed{ \pink{ \rm{distance = speed \times time}}}}}$



${ \implies{ \blue{ \sf{distance = 5 \: \pi \: m \: {sec}^{ - 1} \times 80 \: sec}}}}$



${ \implies{ \sf{ \blue{distance = 400 \: \pi \: m}}}}$



${ \implies{ \sf{ \blue{distance = 400 \times 3.14 \: m}}}}$



${ \implies{ \underline{ \boxed{ \red{ \sf{distance = 1256 \: m}}}}}}$