Solving for Speed: Exploring Velocity and Magnitude

In summary, the conversation discusses the relationship between velocity, speed, and vertical and horizontal components in a skiing scenario. It is clarified that at the bottom of an incline, the magnitude of the velocity is equal to the horizontal component. The picture provided is deemed misleading as it shows a higher position for point #3. The statement of the problem includes a vertical velocity component of 3 m/s for the skiers. However, there is a slight issue with the general model and the gain of vertical velocity without a change in height. It is concluded that the force provided by the skier's muscles causes the addition of the vertical component of velocity, leading to an increase in speed and kinetic energy.
  • #1
Svelte1
9
2
Homework Statement
Okay, I saw a derivation of work equaling a change in kinetic energy and it explicitly states that this is regarding the MAGNITUDE of the velocity vector, however in an example question the formula is used for the horizontal velocity component and I don't understand how this can be done, seems incorrect to me.
Relevant Equations
Work= change in potential energy=change in kinetic energy
https://ibb.co/jG6n0jZ
The 15 is fine as this is clearly his overall magnitude but then v2 is equated to the horizontal velocity rather than the magnitude.

1651154735712.png
 
Physics news on Phys.org
  • #2
At the bottom of the incline the motion is in the horizontal direction and there is no vertical velocity component. This means that the magnitude of the velocity (speed) is the same as the horizontal component.

On edit: In other words (and an equation), the magnitude of the velocity, also known as speed, is$$v=\sqrt{v_x^2+v_y^2}.$$ When ##v_y=0##, ##v=v_x##.
 
Last edited:
  • Like
Likes PeroK and Svelte1
  • #3
kuruman said:
At the bottom of the incline the motion is in the horizontal direction and there is no vertical velocity component. This means that the magnitude of the velocity (speed) is the same as the horizontal component.
I believe that the picture is misleading, as it incorrectly shows position #3 at a higher level, unless the skier is jumping just when arriving at point #2.
 
  • #4
Lnewqban said:
I believe that the picture is misleading, as it incorrectly shows position #3 at a higher level, unless the skier is jumping just when arriving at point #2.
The statement of the problem includes "##\dots~## they jump upward, achieving a vertical velocity component of 3 m/s." You must have missed it.
 
  • #5
kuruman said:
The statement of the problem includes "##\dots~## they jump upward, achieving a vertical velocity component of 3 m/s." You must have missed it.
I indeed missed it!
Thank you.
 
  • Like
Likes dlgoff, berkeman and kuruman
  • #6
Thanks!

If we use the eq. before they jump then we have 24.8 for horizontal and magnitude of velocity, Then as they leave the ramp I can solve for 25 magnitude.

I still have a slight problem with the general model though. If we use the same equation for when they start to jump, but before they change their height, they have gained a vertical velocity, but we end with 24.8 for the magnitude.
 
  • #7
Svelte1 said:
Thanks!

If we use the eq. before they jump then we have 24.8 for horizontal and magnitude of velocity, Then as they leave the ramp I can solve for 25 magnitude.

I still have a slight problem with the general model though. If we use the same equation for when they start to jump, but before they change their height, they have gained a vertical velocity, but we end with 24.8 for the magnitude.
When the jump has just started, the skier’s muscles have added a vertical component to his existing horizontal velocity by pushing against the surface. That increases the speed and hence the kinetic energy. When skied reaches maximum height, the vertical component is zero and the kinetic energy is back to what it was before the jump.
 
  • Like
Likes Svelte1
  • #8
I suppose they can't gain the vertical velocity without a change in their height either, so I was wrong with what I said. Even if its just an almost infinitesimal amount of distance traveled upwards..
 
  • #9
Svelte1 said:
I suppose they can't gain the vertical velocity without a change in their height either, so I was wrong with what I said. Even if its just an almost infinitesimal amount of distance traveled upwards..
You got it backwards. The height does not cause the vertical velocity. The skier cannot gain height unless a force adds to the existing kinetic energy in such a way as to add a vertical velocity component. That force is provided by the skier's muscles pushing against the surface. The cause is the addition of the vertical component of the velocity that increases the speed and hence the kinetic energy. The effect is the gain in height. When the skier reaches maximum height, the work done by gravity on the skier has taken away the energy that the skier's muscles added while executing the jump.
 
  • Like
Likes Lnewqban
  • #10
I just meant he can't have been said to have any vertical velocity until he has gained some vertical distance.
 
  • #11
Svelte1 said:
I just meant he can't have been said to have any vertical velocity until he has gained some vertical distance.
That is true. The velocity is the rate of change of position with respect to time. If the skier is still in contact with the surface and you know that the velocity has acquired a vertical component at that moment, then you can predict that the skier will be off the surface at the next moment. If you don't know that the skier has acquired a vertical component at that given moment and you see that the skier is off the surface at the next moment, you can infer that the skier has acquired a vertical component at the earlier moment. Acquiring a vertical component is a necessary and sufficient condition for the skier to be off the surface.
 
  • Like
Likes Svelte1

1. What is velocity and how is it different from speed?

Velocity is a vector quantity that describes the rate of change of an object's position in a specific direction. It includes both the speed of an object and its direction of motion. Speed, on the other hand, is a scalar quantity that only describes the rate of change of an object's position, regardless of its direction.

2. How do you calculate velocity?

Velocity is calculated by dividing the change in an object's position (displacement) by the time it took for that change to occur. This can be represented by the formula v = Δx/Δt, where v is velocity, Δx is change in position, and Δt is change in time.

3. What are the units of velocity?

The units of velocity depend on the units used for displacement and time. In the SI system, velocity is measured in meters per second (m/s). In the imperial system, it is measured in feet per second (ft/s) or miles per hour (mph).

4. How does velocity relate to acceleration?

Acceleration is the rate of change of an object's velocity. In other words, it describes how quickly an object's velocity is changing. If an object's velocity is increasing, it is said to have positive acceleration. If its velocity is decreasing, it has negative acceleration (also known as deceleration).

5. How is velocity used in real-world applications?

Velocity is used in many real-world applications, such as in sports to measure the speed of athletes or in transportation to determine the speed of vehicles. It is also an important concept in physics, as it is used to describe the motion of objects and predict their future positions.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
640
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
776
  • Introductory Physics Homework Help
Replies
1
Views
251
  • Introductory Physics Homework Help
Replies
5
Views
1K
Replies
20
Views
870
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
711
Replies
1
Views
1K
Back
Top