Question

A car travelling on a straight road in the same direction covers half a distance at a uniform velocity of 80 km/h and the other half of the distance travelling uniformly at 40 km/h. What is the magnitude of average velocity over the total period of motion?

Answer 1

For first half distance=s1=v1t/2=80t/2

For next half distance =s2=v2t/2=40t/2

Average speed= (80t/2)+(40t/2)/(t/2)+(t/2)

= 60km/h

Hence , average velocity is 60km/h

Answer 2

Answer:

The answer is 53.33 km/h

Explanation:
First - Draw a diagram representing almost everything

Here we want average velocity magnitude, for that

Magnitude of the displacement = 2x km

------------------------------------------------------------

by  total time
So,
Avg. velocity magnitude = 2x km/h ÷ t
Now to find time,

Displacement = Velocity / Time
So,
Time = Displacement / Velocity

So displacement by the first half = x/80
Displacement by the second half is = x/40

So total time = x/40 × 2/2 (LCM) + x/80 = 3x/80

Now all we got to do is fill them up :

So, 2x km ÷ 3x ÷ 80
     (Reciprocal)
      2x × 80 ÷ 3 = 160 km/h (NOTE : IT IS km/h)        
∴ 53.33 km/h

This image below may help you

That's it ;)