Home
Class 9
PHYSICS
Usha swims in a 90 m long pool. She cove...

Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.

Text Solution

Verified by Experts

Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in `1 min =0 m `
Average speed `= (" Total distance covered")/("Total time taken") = (180m )/(1 mn)`
`= (180 m)/(1 min ) x (1 min)/(60 s)`
`= 3 ms ^(-1)`
Average velocity `= ("Displacement")/("Total time taken") = (0m)/(60s ) = 0 m s ^(-1)`
The average speed of Usha is `3 ms ^(-1) and ` her average velocity is `0 ms ^(-1).`
Promotional Banner

Topper's Solved these Questions

  • MOTION

    KUMAR PRAKASHAN|Exercise ADDITIONAL NUMERICALS FOR PRACTICE |48 Videos
  • MOTION

    KUMAR PRAKASHAN|Exercise FILL IN THE BLANKS |50 Videos
  • MOTION

    KUMAR PRAKASHAN|Exercise ANSWER THE FOLLOWING |24 Videos
  • GRAVITATION

    KUMAR PRAKASHAN|Exercise ACTIVITY 10.4|2 Videos
  • QUESTION PAPER 01

    KUMAR PRAKASHAN|Exercise SECTION - D|5 Videos

Similar Questions

Explore conceptually related problems

An old person moves on a semi circulasr track of radius 40.0 m during a moirning walk. If he starts at one end of the track and reaches at the other end, find the distance covered and the displacement of the person.

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minute 30 second and then turns around and jogs 100 m back to point in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?

A particle is projected vertically up from the top of a tower with velocity 10 m//s . It reaches the ground in 5s. Find - (a) Height of tower (b) Striking velocity of particle at ground (c ) Distance traversed by particle. (d) Average speed & average velocity of particle.

A person moves on a semicircular track of radius 40 m. If he starts at one end of the track and reaches the other end, find the distance covered and magnitude of displacement of the person.

A car is moving along a straight line, say OP in It move from O to P in 18 s and returns from P to Q in 6 s , where OP=360 m and OQ=240 m. What are the average velocity and average speed of the car in going (a) from O to P ? And (b) from O to P and back to Q ?

A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station . A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min . What is (a) the average speed of the taxt , (b) the magnitude of average velocity ? Are the two equal ?

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field vecE. Due to the force q vecE, its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more second under the influence of this filed. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively,