Is Average Velocity Equal to Instantaneous Velocity for Constant Motion?

AI Thread Summary
When an object's velocity is constant, its average velocity equals its instantaneous velocity at every point along its path. This is because, with constant motion, there are no variations in speed. Therefore, the instantaneous velocity, defined as the velocity at any specific point, matches the average velocity. The discussion confirms that for constant motion, the two velocities are indeed equal. This principle is fundamental in understanding motion in physics.
future_vet
Messages
169
Reaction score
0
Can we say that when the velocity is constant, the average velocity of an object is equal to the instantaneous velocity?

Thanks!
 
Physics news on Phys.org
If the velocity is constant, then at every point on the particle's path, the velocity will be the same. Thus the average velocity will be equal to the velocity at any point on the particle's path. If you are defining the instantaneous velocity to be the velocity at any point, then the answer to your question is yes.
 
Thank you so much, you are very very helpful!
 
You're welcome!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top