Kinematics Equations: Displacement Calculation with a Wheel | Homework Help"

In summary: It comes from the Latin word for "to turn." The problem is trying to make the connection between the equation and what you are seeing in the problem.
  • #1

Homework Statement


A straight line is marked in meters in a field. A student begins at the zero position mark and ends at the 50 position mark. However, the student did not walk along the straight line. He uses a wheel that has a circumference of 1m and finds that it turns 100 times as he walks. What is his displacement?

Homework Equations


  • Vf=Vi+a⋅t
  • Δd=Vi(t)+1/2a⋅t
  • Vf^2=Vi^2+2a⋅Δd
  • Δd=1/2(Vf+Vi)t
Vf= Final Velocity,
Vi=Initial Velocity,
a=Acceleration,
t= Time
Δd= Displacement

The Attempt at a Solution


[/B]
I haven't attempted this because I'm not really sure how to approach it.
 
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  • #2
Here's a start. Can you say what displacement means?
 
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  • #3
DracoMalfoy said:

Homework Statement


A straight line is marked in meters in a field. A student begins at the zero position mark and ends at the 50 position mark. However, the student did not walk along the straight line. He uses a wheel that has a circumference of 1m and finds that it turns 100 times as he walks. What is his displacement?

Homework Equations


  • Vf=Vi+a⋅t
  • Δd=Vi(t)+1/2a⋅t
  • Vf^2=Vi^2+2a⋅Δd
  • Δd=1/2(Vf+Vi)t
Vf= Final Velocity,
Vi=Initial Velocity,
a=Acceleration,
t= Time
Δd= Displacement

The Attempt at a Solution


[/B]
I haven't attempted this because I'm not really sure how to approach it.

Start by ditching all the irrelevant equations that you wrote (some of which are incorrect anyway). Think about the problem before writing anything.
 
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  • #4
Wow. Well... That was a big help..
 
  • #5
DracoMalfoy said:
Wow. Well... That was a big help..

I may be being cynical, but I’m wondering if that was sarcasm. If so, you didn’t think about what he said. I will be more explicit.

This problem does not require any math. Seriously. None. Zero. It is all about the definition of displacement. They want to be sure you understand what displacement is and they want to help you avoid a common misconception of what it is (by tempting you to get it wrong). So, as the previous two posters suggested, go look up the definition of displacement and see if you get the distinction they are trying to make with this problem.
 
  • #6
Cutter Ketch said:
I may be being cynical, but I’m wondering if that was sarcasm. If so, you didn’t think about what he said. I will be more explicit.

This problem does not require any math. Seriously. None. Zero. It is all about the definition of displacement. They want to be sure you understand what displacement is and they want to help you avoid a common misconception of what it is (by tempting you to get it wrong). So, as the previous two posters suggested, go look up the definition of displacement and see if you get the distinction they are trying to make with this problem.
Displacement is final velocity minus initial. Initial is 0 and 50 is final? I guess I wasn't thinking that it would be that obvious and somehow overlooked it or forgot. The wheel thing kind of confused me as well.
 
  • #7
He took a twisty path and his wheel measures 100m. However, the field says he moved 50 yards or something like that. What is his displacement? Hopefully this is clearer.
 
  • #8
Nothing to do with velocity.

Look up the difference between distance and displacement.
 
  • #9
DracoMalfoy said:
Displacement is final velocity minus initial.

No. You really need to crack open your book. It should explain the definition both in words and symbols. Here's another hint: think about the etymology of displacement
 

1. What are the basic kinematics equations?

The basic kinematics equations are:

1. Velocity Equation: v = u + at

2. Displacement Equation: s = ut + 1/2at2

3. Acceleration Equation: v2 = u2 + 2as

4. Time Equation: t = (v-u)/a

5. Displacement Equation (using average velocity): s = (u+v)/2 * t

2. How do I know which equation to use for a given problem?

The equation to use depends on what information is given in the problem. If you are given the initial and final velocities, acceleration, and time, you can use any of the first three equations to solve for the remaining variable. If you are given the initial and final velocities, acceleration, and displacement, you can use the fourth equation to solve for time. If you are given the initial and final velocities, and time, you can use the fifth equation to solve for displacement.

3. Can these equations be used for any type of motion?

These equations are specifically for objects moving with constant acceleration. This means that they can be used for motion with a constant velocity (zero acceleration) or motion with a changing velocity (non-zero acceleration).

4. Do these equations only work for linear motion?

Yes, these equations only apply to linear motion, where an object moves along a straight line. They do not apply to rotational or circular motion.

5. Can these equations be used for objects moving in three dimensions?

Yes, these equations can be used for objects moving in three dimensions, as long as the motion is linear and the acceleration is constant. In this case, the equations will be applied separately to each dimension.

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