Question

a body moving with uniform acceleeation is having velocity 2m/s and 8m/s at t = 1s and t=4s respectively .then average velocity of particle in the time interval t=1s to t=4s is​

Answer 1

average velocity=total distance/total time

=d1+ d2/t1+t2

=v1×t1+v2×t2/t1+t2

=2×1+8×4/4+1

=34/5=6.8m/s

Answer 2

Answer :

Given -

• Velocity at t = 1 s is 2 m/s
• Velocity at t = 4 s is 8 m/s

To Find -

• Average velocity ?

Solution -

We know that, Speed = Distance/Time.

So, Distance = Speed × Time.

Hence, Distance when speed was 2 m/s is : 2 × 1 = 2 m.

And, Distance when speed was 8 m/s is : 8 × 4 = 32 m.

Hence, Total Distance = 34 m.

Total Time = 1 + 4 = 5 s.

We know that, Average Speed = Total Distance/Total Time.

⇒ Avg. Speed = 34/5

⇒ Avg. Speed = 6.8 m/s

Hence, Average Speed of the body is 6.8 m/s.

Additional Information -

• Distance is the total length of the path travelled by the body.
• Displacement is the shortest distance between the initial position and the final position of the body.
• Distance is a scalar quantity whereas displacement is a vector quantity.
• Speed is distance travelled per unit time.
• Velocity is displacement per unit time.
• Speed is a scalar quantity whereas velocity is a vector quantity.