How to Calculate Average Velocity on a Circular Track?

AI Thread Summary
The average velocity of a car traveling on a circular track at a constant speed of 20 m/s is 0 m/s after completing one lap. This is because average velocity is defined as the change in displacement over time, and since the car returns to its starting point, there is no change in displacement. In contrast, the average speed remains 20 m/s, as it measures the total distance traveled divided by time. Understanding the distinction between average velocity and average speed is crucial in solving such problems. Therefore, the key takeaway is that while speed can be constant, average velocity can be zero in circular motion.
cvc121
Messages
61
Reaction score
1

Homework Statement


A car travels on a circular track at 20m/s. The track has a circumference of 2.0km. What is the average velocity of the car over one lap?

Homework Equations

The Attempt at a Solution


Since the car is traveling at a constant speed of 20m/s, wouldn't the average velocity be 20m/s? The answer key says 0.
 
Physics news on Phys.org
Cannot you imagine why 0 is a valid answer?
 
Check the definitions of average velocity and average speed.
 
Oh, my bad. Since the car arrives back at the same location, there is no change in displacement and thus, average velocity is 0.
 
  • Like
Likes PietKuip
cvc121 said:
Oh, my bad. Since the car arrives back at the same location, there is no change in displacement and thus, average velocity is 0.
Yessir! :oldbiggrin:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top