Why Does My Displacement Calculation Exceed the Sprint Distance?

[Balsam]

Homework Statement

I'm just confused about this. I know that displacement=df-di, but using that formula didn't give me the correct displacement in this case.

In a 100m sprint, a runner starts from rest and accellerates to 9.6m/s[W] in 4.2 seconds. The runner runs at a constant velocity for the rest of the race, what is the total time?

I am trying to solve for the time that it takes for the runner to go from their first displacement, when they are accelerating, to the end of the race when they're running at a constant velocity. In a previous calculation, the first displacement is 20m[W]. So, the final displacement- from when the runner is traveling at a constant velocity until they finish the race, should be 80m[W] because the whole sprint it 100m long. But, if you do displacement=d2-d1 and try to solve for d2 that way, you get a displacement of 120m[W]. For my calculations, I plugged in all values in the west directions with negative signs.
Conventions-- West= negative, East= positive

-100=d2-(-20)
-100-20=d2
-120=d2
120m[W]=d2.
How can d2 be 120 m[W] if the sprint itself is only 100m long. Can someone please explain this to me?

Homework Equations

displacement=d2-d1

The Attempt at a Solution

I showed that above, under the problem statement. Can someone tell me where I went wrong? I'm really confused.

 

[J Hann]

Since he's already run -20 shouldn't
d2 = -100 - (-20) ?
or D = d2 - d1 where D is the distance remaining.

Displacement is a vector quantity that represents the change in position of an object from its initial point to its final point. It has both magnitude (how far) and direction (where it moved). To calculate displacement, you need to know the initial position and the final position of the object. Here's how you can find displacement:

1. Identify Initial and Final Positions:
   • Determine the initial position (starting point) of the object. This position is usually represented by a point or coordinates in space.
   • Find the final position (ending point) of the object. Similar to the initial position, the final position is also represented by a point or coordinates.
2. Calculate the Change in Position:
   • To calculate displacement, you subtract the initial position from the final position. This is done vectorially to account for direction.
   • If you have coordinates (x, y, z) for the initial and final positions, you can find the displacement vector (Δx, Δy, Δz) using the following formula:Displacement = (Final Position) - (Initial Position)Δx = x(final) - x(initial)Δy = y(final) - y(initial)Δz = z(final) - z(initial)
3. Determine the Magnitude and Direction:
   • The magnitude of the displacement vector represents the distance between the initial and final positions. You can calculate it using the Pythagorean theorem (for 2D displacement) or the 3D equivalent for three-dimensional displacement.
   • The direction of the displacement vector can be determined by finding the angles or the vector components that specify the direction. This involves trigonometry.

For 2D Displacement:

• Magnitude (D) = √(Δx^2 + Δy^2)
• Direction (θ) = atan(Δy / Δx)

For 3D Displacement:

• Magnitude (D) = √(Δx^2 + Δy^2 + Δz^2)
• Direction can be determined using spherical coordinates or by finding the angles between the displacement vector and each axis.

1. Units:
   • Make sure that the units of your displacement match the units of your initial and final positions. If you are working in meters, your displacement should also be in meters.

Keep in mind that displacement is a vector, so it has both magnitude and direction. In contrast, distance is a scalar that only represents how much ground is covered and doesn't include direction information. Displacement is a more comprehensive measure when you need to know the specific change in an object's position.

 

[Balsam]

Since he's already run -20 shouldn't
d2 = -100 - (-20) ?
or D = d2 - d1 where D is the distance remaining.

Shouldn't you be able to use d=df-di and solve for df by isolating for it. If you do that, you don't get the correct answer.

 

[SteamKing]

Shouldn't you be able to use d=df-di and solve for df by isolating for it. If you do that, you don't get the correct answer.

I think what J Hann is trying to point out is that you are told the runner is running west after he starts the race. The race has a total distance of 100 m, so saying that the runner runs for 120 m is a little ... odd.

If you draw a diagram of the race, making the starting point the origin, this should help you visualize what happens while the runner accelerates and when he crosses the finish line.

 

[CWatters]

But, if you do displacement=d2-d1 and try to solve for d2 that way, you get a displacement of 120m[W]. For my calculations, I plugged in all values in the west directions with negative signs.
Conventions-- West= negative, East= positive

-100=d2-(-20)
-100-20=d2
-120=d2
120m[W]=d2.

I think your equation for the displacement in the second phase should be...

Displacement = Dfinal - Dinitial
= -100 - (-20)
= -80

 

[CWatters]

The problem statement does not appear to specify which direction he is running so why make it harder by choosing a negative direction?

 

[CWatters]

The problem statement does not appear to specify which direction he is running so why make it harder by choosing a negative direction?

Actually is see it says he has a positive Velocity and positive displacement at the end of the acceleration phase.

 

[J Hann]

Using your sign conventions
-100 = d1 + d2 where d1 and d2 are the distances traveled.
Using d2 - d1 you would be referencing coordinates not the distance traveled.

 

[Balsam]

Using your sign conventions
-100 = d1 + d2 where d1 and d2 are the distances traveled.
Using d2 - d1 you would be referencing coordinates not the distance traveled.

Isn't the formula displacement=d2-d1, I subbed in the coordinates for the given points

 

[haruspex]

Isn't the formula displacement=d2-d1, I subbed in the coordinates for the given points

That depends on what d1 and d2 stand for. If d1 is an initial displacement and d2 is the final displacement then the extra displacement is indeed d2-d1. To apply your formula, you need to set d2=100, as that is the final displacement. With d1=20, the extra displacement will be the length at constant velocity.

 

[ehild]

Homework Statement

I'm just confused about this. I know that displacement=df-di, but using that formula didn't give me the correct displacement in this case.
.

You need the clear definition of displacement. It can not be defined by itself.
Displacement is the change of position. If an object moves along the x-axis from position xi to the final position xf, then the displacement is d = xf-xi.
If the object moves from xi to x1 and then to x1 to x2, then the first displacement is d1=x1-xi the second displacement is d2=xf-x1, and the total displacement is d=xf-xi=d1+d2.
In this case, you took xi =0 and xf=-100m.The total displacement is d=-100m. You calculated the first displacement, it was -20m. What is the second displacement?
Note that the velocity is also negative if you take westward direction negative.

 

[CWatters]

+1

That's why in an earlier post I used "final" and "initial". I guess I should have spelled it out a bit more..

Displacement = Dfinal - Dinitial
= -100 - (-20)
= -80