Question

an athelete completes one round on a circular diameter 200m in 40 seconds . what would be distance covered and displacement at the end of 2 minutes 20 seconds ?​

Answer 1

Answer:-

Displacement is the shortest distance possible between initial and final position.

As the athlete is in circular motion, displacement is zero. It is because, the initial and final position is the same.

Distance covered in 1 round = Circumference of the circle

Circumference of the circle = 2πr

Diameter = 200 m

Radius = 100 m

Distance covered in 1 round[ 40 sec ] = 2 × 22/7 × 100 = 628.57 m

Distance covered in 1 sec = 628.57 ÷ 40 = 15.71 m

2min 20 sec = 140 sec

Distance covered in 140 sec = 15.71 × 140 = 2199.4 m.

Answer 2

☯ Given :

Athelete completes one round on a circular diameter 200m in 40 seconds.

$\rule{200}{1}$

★ To Find :

We have to find distance covered and displacement at the end of 2 minutes 20 seconds ?

$\rule{200}{1}$

★ Solution :

Distance covered by athelete in one round will be equl to circumference of circle as athlete move arounds the circular path.

We know that,

$\Large{\implies{\boxed{\boxed{\sf{Circumference = 2 \pi r}}}}}$

★ Putting Values ★

$\small{\sf{\dashrightarrow Distance \: covered \: in \: one \: round = 2 \times \frac{22}{7} \times 100}} \\ \\ \small{\sf{\dashrightarrow Distance \: covered \: in \: one \: round = 2 \times \frac{2200}{7}}} \\ \\ \small{\sf{\dashrightarrow Distance \: covered \: in \: one \: round = \frac{4400}{7}}} \\ \\ \small{\sf{\dashrightarrow Distance \: covered \: in \: one \: round = 628.57 \: m}} \\ \\ \small{\implies{\boxed{\boxed{\sf{Distance \: covered \: in \: one \: round = 628.57 \: m}}}}}$

$\rule{200}{1}$

Now, we will find distance covered in one second by deviding distance covered in one round and 40 seconds.

$\small{\sf{\dashrightarrow Distance \: covered \: in \: one \: second = \frac{628.57}{40}}} \\ \\ \small{\sf{\dashrightarrow Distance \: covered \: in \: one \: second = 15.71 m}}\\ \\ \small{\implies{\boxed{\boxed{\sf{Distance \: covered \: in \: 1 \: seconds = 15.71 \: m}}}}}$

$\rule{150}{2}$

Time taken = 2 mins 20 seconds = 140 seconds

→ Multiply distance covered in one second by 140 seconds.

$\small{\sf{\dashrightarrow Distance \: covered \: in \: 140 \: seconds = 15.71 \times 140}} \\ \\ \small{\sf{\dashrightarrow Distance \: covered \: in \: 140 \: seconds = 2199.4 \: m}} \\ \\ \small{\implies{\boxed{\boxed{\sf{Distance \: covered \: in \: 140 \: seconds = 2199.4 \: m}}}}}$