A player completes a circular path of radius r in $40 s$ . At the end of $2$ $m i n u t e s$ $20$ $s e c o n d s$ , its displacement will be

2 r

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 π r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7 π r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Zero

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2 r$
The player completes one complete revolution in $40 s$ .
Time taken $t = 2 m i n 20 s e c = 2 \times 60 + 20 = 140 s$
Thus in this time, the player completes $\frac{140}{40} = 3.5$ revolutions
After this time, he reaches at the diametrically opposite end. Thus his displacement is equal to the diameter of the circular path i.e. $2 r$

Suggest Corrections

Similar questions

An athlete completes one round of a circular track of radius R in 40 sec . What will be his displacement at the end of 2 minutes 20seconds?

A particle is moving along a circle of radius R such that it completes one revolution in 40 seconds. What will be the displacement after 2 minutes 20 seconds?

Q. A person completes a round of circular track of radius

R

40 s

. At the end of

2 minutes

, displacement will be

Q. An athlete completes one round of a circular track of radius

R

40

sec. What will be the magnitude of displacement at the end of

2

min

20

sec?

Q. Question 1
An athlete completes one round of a circular track of diameter 200 m in 40 seconds. What will be the distance covered and the displacement at the end of 2 minutes 20 seconds ?

Join BYJU'S Learning Program

A player completes a circular path of radius r in 40s. At the end of 2 minutes 20 seconds, its displacement will be

A player completes a circular path of radius r in $40 s$ . At the end of $2$ $m i n u t e s$ $20$ $s e c o n d s$ , its displacement will be