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

Answer

Change in velocity

$dv=\sqrt{2v^2}=2.12m/s$

Change in time

$dt=T/4=3.15s$

Average accekeration

$a=\frac{dv}{dt}\\=\frac{2.12}{3.15}\\=0.67m/sec^2$