A car is moving along a straight line, say OP in Fig. It moves from O to P in 18 s and returns from P to Q in 6.0 s. What are the average velocity and average speed of the car in going (a) from O to P ? and (b) from O to P and back to Q ?

A car is moving along a straight line, say OP in Fig. It moves from O

1.	A car is moving along a straight line, say OP in Fig. It moves from O to P in 18 s and returns from P to Q in 6.0 s. What are the average velocity and average speed of the car in going (a) from O to P ? and (b) from O to P and back to Q ?
Answer» Solution :(a)AVERAGE velocity = `("Displacement")/("TIME internal")` `bar(upsilon)=(+360m)/(18S)=+20 ms^(-1)` Average speed `=("Path length")/("Time interval")` `=(360m)/(18s)=20 ms^(-1)` Thus, in this case the average speed is equal to the magnitude of the average velocity. (b) In this case, Average velocity `=("Displacement")/("Time interval")=(+240m)/((18+6.0)s)` `=+10 ms^(-1)` Average speed `=("Path length")/("Time interval")=(OP+PQ)/(Delta t)` `=((360+120)m)/(24s)=20 ms^(-1)` Thus, in this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves change in direction so that the path length is greater than the magnitude of displacement. This shows that speed is, in general, greater than the magnitude of the velocity.