Average Speed - isn't this average velocity?

AI Thread Summary
The discussion revolves around calculating the average speed and average velocity of a billiard ball's motion. The average speed is determined by adding the total distance traveled (0.44m + 0.88m + 0.12m) and dividing by the total time (2.4s), resulting in 0.60 m/s. In contrast, average velocity considers the net displacement from the initial to final position, which leads to confusion as it yields a different value. Participants clarify that average speed is a scalar quantity based on total distance, while average velocity is a vector based on displacement. Additionally, guidance is sought on calculating the final position of the ball, which is determined through vector addition rather than kinematic equations.
KaseyK
Messages
8
Reaction score
0
A billiard ball travels 0.44m from its original position, bounces off another ball and travels 0.88m [N], then bounces off the edge of the billiard table coming to rest 0.12m from that edge. The entire motion is one-dimentional and takes 2.4s. (a) calculate the average speed of the ball.

Okay the answer at the back of the book is 0.60 m/s. Firstly, I thought that 'average speed' is actually 'average velocity'.

If that is the case then average velocity is delta d / t. If I take 0.12(d2) - 0 (d1) / 2.4 I get 0.05 m/s. Which according to the book is incorrect. The only way that I get the correct answer if it I add up all the distances (0+0.44+0.88+0.12) and divide by the time (2.4) = 0.60 m/s.

I'm confused on why I should be adding up all the distances instead of taking the final distance minus the original distance, divided by the time?

What am I missing here?
Thanks in advance.
 
Physics news on Phys.org
KaseyK said:
A billiard ball travels 0.44m from its original position, bounces off another ball and travels 0.88m [N], then bounces off the edge of the billiard table coming to rest 0.12m from that edge. The entire motion is one-dimentional and takes 2.4s. (a) calculate the average speed of the ball.

Okay the answer at the back of the book is 0.60 m/s. Firstly, I thought that 'average speed' is actually 'average velocity'.

If that is the case then average velocity is delta d / t. If I take 0.12(d2) - 0 (d1) / 2.4 I get 0.05 m/s. Which according to the book is incorrect. The only way that I get the correct answer if it I add up all the distances (0+0.44+0.88+0.12) and divide by the time (2.4) = 0.60 m/s.

I'm confused on why I should be adding up all the distances instead of taking the final distance minus the original distance, divided by the time?

What am I missing here?
Thanks in advance.


Speed is the instantaneous rate of displacement. It is a scalar of the velocity vector.

If you figure average speed, then you need total distance traveled.

If it would be average velocity then it would be just the displacement from initial location to final.
 
Thank you very much. That is so much more clearer.

I did have an additional question if you don't mind. The (b) part asks me to calculate the final position of the ball. The answer is 0.32m [N].

I was trying to use one of the various kinematic equations however I feel like I'm missing information.

For example I don't have the acceleration so anything with acceleration in it I have to avoid. That leaves only one equation: delta d = 1/2(v2+v1)delta t

V1 is zero but I don't have V2 so I can't use this equation either.

Are you guys able to offer some advise or tips for solving this? I certainly don't expect you to do this for me I'm just looking for some guideance.

Thanks so much.
Kasey
 
KaseyK said:
Thank you very much. That is so much more clearer.

I did have an additional question if you don't mind. The (b) part asks me to calculate the final position of the ball. The answer is 0.32m [N].

I was trying to use one of the various kinematic equations however I feel like I'm missing information.

For example I don't have the acceleration so anything with acceleration in it I have to avoid. That leaves only one equation: delta d = 1/2(v2+v1)delta t

V1 is zero but I don't have V2 so I can't use this equation either.

Are you guys able to offer some advise or tips for solving this? I certainly don't expect you to do this for me I'm just looking for some guideance.

Thanks so much.
Kasey

The problem is not one of equations. It's vector addition.

.44 S, .88 N, .12 S

.44 S = - .44 N

.88 - .44 - .12 = .32 N
 
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 hanged 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