Question

a particle moves one fourth of a distance with a velocity 2m/sec the remaining part it has covered by 4m/sec for the fisrt half time and by 6m/sec for next half find average speed​

Answer 1

Answer :

Given -

• One fourth of the Distance is travelled at a velocity of 2m/sec
• From the remaining Distance, half of it was covered at a velocity of 4m/sec
• And from the remaining distance the other half was covered at a velocity of 6m/sec

To prove -

• The Average speed ?

Answer -

Let us assume the total Distance be x.

Now, We know that, Speed = Distance/Time.

We have been given the Distance and speed/velocity of the motion.

Hence, We need to find the time taken.

➜ Time = Distance/Speed

So, for the first case :

• Distance = d/4 m
• Velocity = 2 m/sec

T1 = (d/4)/2 = d/8 seconds

And, for the second case :

• Distance = (3d/4)/2 = 3d/8 m
• Velocity = 4 m/sec

T2 = (3d/8)/4 = 3d/32 seconds

And, for the third case :

• Distance = 3d/8 m
• Velocity = 6 m/sec

T3 = (3d/8)/6 = 3d/48 seconds

AVERAGE SPEED : - DISTANCE/TIME

➸ d/(d/8 + 3d/32 + 3d/48)

➸ d/(12d + 9d + 6d)/96

➸ 96d/(12d + 9d + 6d)

➸ 96d/27d

➸ 96/27

➸ 3.55 m/sec