Question

A car is moving along a straight line path OP from O to P in 18s and returns to Q in

6s. What are the average velocity and average speed of the car in moving from

a) O to P and

b) O to P and back to Q

Answer 1

a) average speed = 4/18 = 0.222km/s
average velocity =4/18 = 0.222 km/s from O to P


b)average speed =5/18 = 0.277km/s
average displacement = 3/18 = 0.166km/s from O to Q

Answer 2

${\huge{\underline{\underline{\mathfrak{\red{♡ANSWER♡}}}}}}$

a)

$\purple{\bold{\underline{\underline{Average velocity}}}}$

= $\frac{Displacement}{Time interval}$

= v = $\frac{+360m}{18s}$ =20m${s}^-1$.

$\red{\bold{\underline{\underline{Average speed}}}}$

=$\frac{Pathlength}{Time interval}$

= $\frac{360m}{18s}$ = 20m${s}^-1$.

✴.Thus,in this case the average speed is equal to the magnitude of the average velocity.

b)

$\blue{\bold{\underline{\underline{Average Velocity}}}}$

= $\frac{Displacement}{Time interval}$.

$\frac{+240m}{18+6.0}s$

= ${+10ms}^-1$

$\pink{\bold{\underline{\underline{Average speed}}}}$

=$\frac{pathlength}{Time interval}$=$\frac{OP+PQ}{\delta t}$.

=$\frac{(360+120)m}{24s}$ =${20ms}^-1$