Question

particle has a displacement of 40m along OX in the first 5s and 30m perpendicular to OX in

the next 5s. The magnitude of average velocity of the particle during the time internal of 10s is

a) 6 ms–1 b) 7 ms–1 c) 8 ms–1 d) 5 ms–1 [ ​

Answer 1

Given :

Particle has a displacement of 40m along OX in the first 5s and 30m perpendicular to OX in the next 5s.

To Find :

Average velocity of particle

Solution :

During First 5 seconds

Distance travelled = 40 m

time taken = 5 seconds

V1 = distance/time = 40/5 = 8 m/s

During Other 5 seconds

Distance travelled = 30 m

time taken = 5 seconds

V2 = distance/time = 30/5 = 6 m/s

Average velocity = ( V1 + V2 )/2

Average velocity = (8+6)/2

Average velocity = 7 m/s

or

Average velocity = Total distance travelled/ total time taken

Average velocity = ( 40+30 )/( 5+5 )

Average velocity = 70/10

Average velocity = 7 m/s

Hence , average velocity of particle is 7 m/s

So correct option is B

Answer 2

Answer:

option b is the correct answer

Explanation:

hope it was helpfull thank you