Relative motion on a bike

  • #26
RUber
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Yes
That's not right.
You cannot directly subtract the magnitudes of the velocity from each other. This only works when the velocities are pointing in the same direction.

The total velocity of the apple when it starts its journey is 8.731m/sec. And it is pointed at a 20 degree elevation, right toward the trash can. In the x-y plane, it will have a straight line course, in the plane drawn between z and the direction of travel, it will have a standard parabolic ballistic trajectory.

You can answer part C based on the initial angle and initial velocity.
For part B, you need to break the velocity into components, x- component, y-component, z-component. Double check your components with
##v_x ^2 + v_y^2 + v_z^2 = V^2 ##.
Once you have components, you can directly subtract the velocity from the bike -- component-wise. That way you are only dealing with velocities in the same directions.

Review the posts in this thread. There have been at least a couple people who have posted the correct answers...and quite a few who have posted incorrect methods.
 
  • #27
That will give you the magnitude of your velocity (i.e. speed).
Most vectors are expressed in their component form, which you already have.
So, in b) if I have vectors Vi=8,73, Vb(bike)=5,6 (that is m/s from 20 km/h) and Va...Va=3,13.
To find initial velocityof the throw relative to me on the bike sqrVb+sqrVa=sqrx, X=6,4m/s.
Relative to me on the bike, displacement is 3 meters...then I can find time t=3/6,4=0.5 sec
The speed will be: dist. traveled by the apple (5m)/time (0,5s)=10 m/s
 
  • #28
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So, in b) if I have vectors Vi=8,73, Vb(bike)=5,6 (that is m/s from 20 km/h) and Va...Va=3,13.
To find initial velocityof the throw relative to me on the bike sqrVb+sqrVa=sqrx, X=6,4m/s.
Relative to me on the bike, displacement is 3 meters...then I can find time t=3/6,4=0.5 sec
The speed will be: dist. traveled by the apple (5m)/time (0,5s)=10 m/s
What is Va? Are you still subtracting the velocity of the bike directly from the velocity of the apple? How can the speed you calculated at the end be more than the speed you started with of 8.73? That is not right.

First, break the initial velocity into its components.

8.731 * sin(20) is your vertical component
8.731 * cos(20) is your flat component.
3/5 ( 8.731 * cos(20) ) is the portion of the flat component moving to the *edit* right.
4/5 ( 8.731 * cos(20) ) is the portion of the flat component moving forward.

Then, and only then, can you subtract the velocity of the bike. And, you only subtract the velocity of the bike from the component pointing in the same direction as the bike.
Once you have done that, you have your answer...in vector component form.

If you want to see the speed, then take sqrt ( [8.731 * sin(20)]^2 + [3/5 ( 8.731 * cos(20) ) ]^2 + [ 4/5 ( 8.731 * cos(20) ) - Vbike ] ^2 ).
That is the speed in the direction of the throw.
 
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  • #29
What is Va? Are you still subtracting the velocity of the bike directly from the velocity of the apple? How can the speed you calculated at the end be more than the speed you started with of 8.73? That is not right.

First, break the initial velocity into its components.

8.731 * sin(20) is your vertical component
8.731 * cos(20) is your flat component.
3/5 ( 8.731 * cos(20) ) is the portion of the flat component moving to the left.
4/5 ( 8.731 * cos(20) ) is the portion of the flat component moving forward.

Then, and only then, can you subtract the velocity of the bike. And, you only subtract the velocity of the bike from the component pointing in the same direction as the bike.
Once you have done that, you have your answer...in vector component form.

If you want to see the speed, then take sqrt ( [8.731 * sin(20)]^2 + [3/5 ( 8.731 * cos(20) ) ]^2 + [ 4/5 ( 8.731 * cos(20) ) - Vbike ] ^2 ).
That is the speed in the direction of the throw.
Va is a relative velocity. http://www.physicstutorials.org/home/mechanics/1d-kinematics/relative-motion
 
  • #30
I just do not get why is there portions moving to the left, as you say....
 
  • #32
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I just do not get why is there portions moving to the left, as you say....
Edited, that was moving to the right in the diagram.
 
  • #33
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What is Va? Are you still subtracting the velocity of the bike directly from the velocity of the apple? How can the speed you calculated at the end be more than the speed you started with of 8.73? That is not right.

First, break the initial velocity into its components.

8.731 * sin(20) is your vertical component
8.731 * cos(20) is your flat component.
3/5 ( 8.731 * cos(20) ) is the portion of the flat component moving to the *edit* right.
4/5 ( 8.731 * cos(20) ) is the portion of the flat component moving forward.

Then, and only then, can you subtract the velocity of the bike. And, you only subtract the velocity of the bike from the component pointing in the same direction as the bike.
Once you have done that, you have your answer...in vector component form.

If you want to see the speed, then take sqrt ( [8.731 * sin(20)]^2 + [3/5 ( 8.731 * cos(20) ) ]^2 + [ 4/5 ( 8.731 * cos(20) ) - Vbike ] ^2 ).
That is the speed in the direction of the throw.

where do you get the angle 20 from?
 
  • #34
Thank you for clarifying.
In any case, you did not add your vectors properly.
Va + Vb = Vi, right? And Vi has vertical, right, and forward components. Vb only has a forward component.
So Va will have the same vertical and right component as Vi. Va's forward component will be Vi(forward) - Vb.
Yes, if you look at the diagram as at is in the problem...origin is the bike, x to the right, y -up. The Vi forward component is equal 8,73...no?
 
  • #35
If we look at the situation as flat diagram, vi(flying apple) is 8,73. Vb is 5,6. Given that relative velocity = v object (Apple)-v observer(me on the bike). I find angles irrelative.
 
  • #36
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Yes, if you look at the diagram as at is in the problem...origin is the bike, x to the right, y -up. The Vi forward component is equal 8,73...no?
No. 8.73m/sec was the total magnitude of the initial velocity. Each component, accordingly would be less than that.
 
  • #37
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If we look at the situation as flat diagram, vi(flying apple) is 8,73. Vb is 5,6. Given that relative velocity = v object (Apple)-v observer(me on the bike). I find angles irrelative.
Without angles, how do you properly add the vectors?

The bike is not travelling in the same direction as the apple, so subtraction of magnitudes does not work.
 
  • #38
Without angles, how do you properly add the vectors?

The bike is not travelling in the same direction as the apple, so subtraction of magnitudes does not work.
For Vi= xi+yj
 
  • #41
What about Vzi?
I am looking at situation schematicly(flat). All vectors then have only two components.
 
  • #42
If I'm on the bike, I do not see the Vzi. Apple just moves away. I do not know what to do anymore.
 
  • #43
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Okay. Then you need to flatten the initial velocity vector, which has 3 dimensions. You can do this by multiplying by the cosine of the elevation angle.
 
  • #44
Okay. Then you need to flatten the initial velocity vector, which has 3 dimensions. You can do this by multiplying by the cosine of the elevation angle.
Then I get 8,2. 8,2-5,6=2,6 is then relative velocity.
 
  • #45
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No. The bike is not moving in the direction of the trash can.
The bike is moving forward.
The forward velocity is 4/5 of the flat velocity.
 
  • #46
No. The bike is not moving in the direction of the trash can.
The bike is moving forward.
The forward velocity is 4/5 of the flat velocity.
I will look at it tomorrow with fresh eyes. Do not want to bother you anymore here. Thank you.
 
  • #47
No. The bike is not moving in the direction of the trash can.
The bike is moving forward.
The forward velocity is 4/5 of the flat velocity.
RUber, be proud. You are great! I got it, was too tired to understand.
 
  • #48
RUber
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RUber, be proud. You are great! I got it, was too tired to understand.
Thanks. Glad it made more sense after some sleep.
 

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