Answer to Question #174244 in Mechanics | Relativity for Aya Haj Nassan

Question

A model car is moving around a circular track of radius 3.0 m at a speed of 1.5 m s⁻¹ as shown in the figure. It takes the car 12.6 s to complete one full lap.

What is the magnitude of the average velocity of the car (a) between point A and point B? (in m s⁻¹ to 2 s.f)

NOTE: The motion of the car is NOT an example of constant acceleration.

Answer 1

As we know that average velocity is given by,

$\boxed{v_{avg}=<v>={Total.displacement \over total .time .taken}={r_f-r_i\over \Delta t}}$

as we can see in the given figure that points A and B are on opposite sides of diameter,

thus velocity vector will rotate 180^0.

and from above equation we can see that average velocity involve vector some of initial and final position .

..thus this sum will be 0 ...

and we will get....

$\boxed{<v_>={r_f-r_i\over \Delta t}=0 m/s}$