Uniformly Accelerated Motion(related to direction)

  • #1

Main Question or Discussion Point

I have some problem with velocity and gravitational acceleration.
Here is my question and answer as it is in my text book, which I could not understand:

A runner make one lap around a 200m track in a time of 25s. What were the runner's (a) average speed and (b) average velocity?
(a)From the definition,
Average speed=[tex]\frac{distance traveled}{time taken}[/tex]
Average speed=[tex]\frac{200m}{25s}[/tex]=8.0m/s
(b)Because the run ended at the starting point, the displacement vector from starting point to end point has zero length. Since [tex]\overline{v}[/tex]=[tex]\frac{\overline{s}}{t}[/tex],
[tex]\overline{v}[/tex]=[tex]\frac{0m}{25s}[/tex]=0m/s

Now, here is my that question in which I have problem:

A ball that is thrown vertically upward on the Moon returns to its starting point in 4.0s. The acceleration due to gravity there is 1.60m/s2downward. Find the ball's original speed.
Let us take up as positive. For the trip from beginning to end, y=0(it ends at the same level it started at), a=-1.60m/s2, t=4.0s. We use y=viyt+½at2 to find
0=viy(4.0s)+½(-1.60m/s2)(4.0s)2
from which viy=3.2m/s.

As you see in 1st question that when calculating velocity then we have distance is 0 while in the 2nd question it is clearly said to calculate ball's original speed.
If we are calculating ball's original speed then why we took y=0? While we are calculating speed in which direction doesn't matter.

Can anyone explain this for me???
 
Last edited:

Answers and Replies

  • #2
Doc Al
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As you see in 1st question that when calculating velocity then we have distance is 0 while in the 2nd question it is clearly said to calculate ball's original speed.
The first question asks for average velocity while the second question (with the ball) asks for the initial velocity.
If we are calculating ball's original speed then why we took y=0?
The ball returns to the starting point, so in your formula y must equal 0.

A more complete formula would be:
y= yi +viyt+½at2

The reference point for measuring vertical position is arbitrary. What's important is that the ball returns to the starting point, so y = yi.

I'm not quite sure if I addressed your issue, so you may need to rephrase your question and try again.
 
  • #3
The first question asks for average velocity while the second question (with the ball) asks for the initial velocity.

The ball returns to the starting point, so in your formula y must equal 0.
How could you say they are asking for initial velocity while they clearly asked for original SPEED?
Help me in understanding to this concept.
 
  • #4
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And what exactly is the difference between velocity and speed?
 
  • #5
And what exactly is the difference between velocity and speed?
There is difference between velocity and speed.

Speed can never be negative because it is the magnitude of velocity and also it is a scalar quantity having no particular direction.

While velocity can be negative due to its direction.

Yes, there is a difference between velocity and speed.
 
  • #6
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So velocity is a vector quantity. Never really made that distinction, though it appears to be valid.

And if we deal in vector quantities, that thing where you put the 0 - the left side of the equation is the displacement vector.
So ask yourself - where is the ball at the end of the journey in relation to the beginning?

The scalar component of that vector would represent the distance between the starting and the ending positions of the ball.
And again - what is that distance?
 
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  • #7
Doc Al
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How could you say they are asking for initial velocity while they clearly asked for original SPEED?
True, they just ask for the initial speed, not velocity. (But I assume you know that the ball was thrown upward.)
Help me in understanding to this concept.
Is there something you don't understand? I thought your issue was instantaneous versus average quantities.
 
  • #8
Rap
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When you say y=0, y is a position, not a direction. You solved the equation correctly. Viy=3.2 m/s and Vix=0, and speed is square root of Vix^2+Viy^2 = 3.2 m/s
 
  • #9
This reply to three of you:
Matrix, Doc Al & Rap
Ok so you people are saying that they ask to find initial velocity by saying original speed?
Well, ok. I consider you people are right. Well thanks.
I close this topic here.
And I have another problem also related to this topic, which will be discuss in new thread. I hope you people again help me. Thanks everyone who helped me.
 
  • #10
Rap
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This reply to three of you:

Ok so you people are saying that they ask to find initial velocity by saying original speed?
No, you are right. They ask for initial speed and the initial speed is 3.2 m/s. The initial velocity is 3.2 m/s upward.
 
  • #11
No, you are right. They ask for initial speed and the initial speed is 3.2 m/s. The initial velocity is 3.2 m/s upward.
Ok. Ok I am 100% agree with your statement. No doubt I understand you.
But if they are asking for speed whatever it is, either it is initial or final or original, nothing matter.
Matter is why they take y=0? While this statement(y=0) is used related to direction and in speed(a scalar quantity) no direction is considered.
I still don't get that point for which I started this thread.

What can I do?
I think I should finish this thread.
 
  • #12
Doc Al
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Matter is why they take y=0? While this statement(y=0) is used related to direction and in speed(a scalar quantity) no direction is considered.
Y is the vertical position of the ball. Y = 0 just means that it's back to its original position.
 
  • #13
Y is the vertical position of the ball. Y = 0 just means that it's back to its original position.
Ok. If say this then listen me dear friend.
Look at my 1st post of this thread. Now, look through my 1st question in that post.
I have calculated average speed. But when I calculated average velocity then I considered y=0(because starting and ending is at same level).
So, then why in finding original SPEED we take y=0?
As you clearly see in that post. Please try to understand me what am I saying.
 
  • #14
Doc Al
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Ok. If say this then listen me dear friend.
Look at my 1st post of this thread. Now, look through my 1st question in that post.
I have calculated average speed. But when I calculated average velocity then I considered y=0(because starting and ending is at same level).
So, then why in finding original SPEED we take y=0?
As you clearly see in that post. Please try to understand me what am I saying.
In your first problem you are calculating AVERAGE quantities. The displacement was zero, so the average velocity was zero.

The second problem is totally different. You are not calculating the average speed during the rise and fall of the ball. (That would be total distance traveled over time.) You are calculating the instantaneous speed at t = 0.
 
  • #15
sophiecentaur
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@shayaan_musta
Are you having difficulty in understanding and using the well known Equations of Motion (aka SUVAT)?
 
  • #16
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Matter is why they take y=0? While this statement(y=0) is used related to direction and in speed(a scalar quantity) no direction is considered.
I still don't get that point for which I started this thread.
By specifying y = 0, they want you to workout the speed at which the ball is thrown.

In other words, before acceleration due to gravity takes effect. At any y > 0, acceleration due to gravity can be considered to be acting on the ball and so it will be accelerating / decelerating.

Speed at y = 0 is original speed which, given we know the direction it is thrown is the initial velocity.

You do know the direction it is thrown (upward) as it's in the question.
 
  • #17
In your first problem you are calculating AVERAGE quantities. The displacement was zero, so the average velocity was zero.

The second problem is totally different. You are not calculating the average speed during the rise and fall of the ball. (That would be total distance traveled over time.) You are calculating the instantaneous speed at t = 0.
Now that's the point. Now I got my destination. Thanks a lot for helping me.

Are you having difficulty in understanding and using the well known Equations of Motion
No no. I better understand all of them. But if I say yes so can you help me?

By specifying y = 0, they want you to workout the speed at which the ball is thrown.

In other words, before acceleration due to gravity takes effect. At any y > 0, acceleration due to gravity can be considered to be acting on the ball and so it will be accelerating / decelerating.

Speed at y = 0 is original speed which, given we know the direction it is thrown is the initial velocity.

You do know the direction it is thrown (upward) as it's in the question.
Thanks for contribution in this thread. I have understood.

And please contribute in my next thread.
You people are great.
Thanks once again.
 

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