"The magnitude of average velocity is equal to average speed". This statement is not always correct and not always incorrect. Explain with example,

Explanation

Supose, a car taken 18 s to move from O (0) to P (360. Noe in next 6s, it returns from P to Q (240) then by calculating average velocity and average speed the answer of this question can be explained.
For motion for O to P,
Average speed `= (OP)/(t _(1)) = (+ 360)/(18) = 20 ms ^(-1)`
Aberage velocity `= ("Position of P-Position of Q")/(t _(1))`
`= (360-0)/(18)= (360)/(18)=20 ms ^(-1)`
Here, both are equal,
For motion from P to Q,
Avewrage speed `= (OP +PQ)(t _(2)) -= (360)/(t_(1) + 6)`
`[ because PQ =360 - 240 = 120]`
` = (480 )/(18 + 6) = (480)/(24) = 20 ms ^(-1)`
Average velocity `= (Position Q-Position Of O")/(t _(2))`
`= (240)/(24)= 10 ms ^(-1)`
Here, both are not equal .