# Average Speed - isn't this average velocity?

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?

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LowlyPion
Homework Helper
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?

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

LowlyPion
Homework Helper
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