Question

06 An athlete completes one round of a circular track of diameter 100 m in 1 minute. What will
be the magnitude of displacement and distance covered after 2 minutes​

Answer 1

Given :

▪ Diameter of circular path = 100m

▪ Time taken by athlete to complete one round = 1min

To Find :

▪ Magnitude of displacement and distance covered by athlete after 2 minutes.

Concept :

✴ Distance is defined as the actual length of total path which is covered by the moving object.

✴ It is a scalar quantity.

✴ It has only magnitude.

✴ It can't be negative or zero.

✒ Displacement is defined as the shortest distance b/w initial and final point of the whole journey covered by the moving body.

✒ It is a vector quantity.

✒ It has both, magnitude as well as direction.

✒ It can be positive, negative and zero.

____________________________

▪ Athelete takes one minute to complete one round.

▪ Therefore, athelete will complete two rounds after two minutes.

▪ Radius of circular path = 100/2 = 50m

▶ Distance (D) :

✏ D = 2×2πR

✏ D = 2×2×3.14×50

✏ D = 628m

▶ Displacement (S) :

→ S = Δ(position)

[Note : Δ(position) = 0 because athelete returns back to the initial position after 2 minutes]

→ S = zero

Answer 2

Given:

Diameter of Track, D= 100 m

Time taken by athlete to complete 1 round of track, t= 1 min

To Find:

Magnitude of displacement and distance covered after 2 minutes​

Solution:

We know that,

• $\pink{\boxed{\bf{Radius=\frac{Diameter}{2}}}}$

• $\purple{\boxed{\bf{Circumference=2\pi r}}}$

where r is radius

• Distance is the length of actual path covered by body

• Displacement is the distance between starting and final position

$\rule{190}{1}$

Let the radius of circular track be r

So,

$\longrightarrow\rm{r=\dfrac{D}{2}}$

$\longrightarrow\rm{r=\dfrac{100}{2}=50\:m}$

$\rule{190}{1}$

Let the circumference of circular track be C

So,

$\longrightarrow\rm{C=2\pi r}$

$\longrightarrow\rm{C=2\times\dfrac{22}{7}\times50}$

$\longrightarrow\rm{C=\dfrac{2200}{7}m}$

$\rule{190}{1}$

If athlete completes 1 round of circular track in 1 min, then he will complete  

2 rounds of circular track in 2 min

Let the distance covered by athlete be d and magnitude of displacement of athlete be d'

So,

$\longrightarrow\rm{d=2C}$

$\longrightarrow\rm{d=2\times\dfrac{2200}{7}}$

$\longrightarrow\rm{d=\dfrac{4400}{7}}$

$\longrightarrow\rm\green{d=628.57\:m}$

Also,

$\longrightarrow\rm\green{d'=0\:m}$

Since, after covering 2 rounds, athlete returns to his starting position

Hence, magnitude of displacement is 0 m and distance covered by athlete is 628.57 m.