What Is the X Coordinate of the Particle?

[Meadman23]

Homework Statement

The graph below represents the time versus velocity of a particle. The x coordinate of the particle at 30 seconds is 20 feet.

What is the x coordinate of the particle at 50 seconds?

Homework Equations

These are the equations I think could be used to solve this problem; the class hadn't been exposed to much else at this point.

Eq. 1 v = v₀ + at

Eq. 2 x - x₀ = v₀t + 1/2at²

Eq. 3 v² = v²₀ + 2a(x - x₀)

Eq. 4 x - x₀ = 1/2(v₀ + v)t

Eq. 5 x - x₀ = vt + 1/2at²

The Attempt at a Solution

The teacher told us that the answer to this problem is -180 but definitely NOT -40.

My method went like this:

I assumed when the problem asked me for the x coordinate, it meant x as in position, but not x as in the x and y-axis on the graph below. From that, I set out to solve this problem using equation 2 and setting it up for final position like so:

x = v₀t + 1/2at² + x₀

then since I knew I didn't have the acceleration for this problem, I tried to find it:

a = -40 - 20 / 50 - 30 = -3ft/s

once I got the acceleration of -3ft/s I added my known values to the equation I setup:

x = 20*50 + (1/2*-3)50² + 20

and then I got:

x = 1000 + (-3750) + 20 = -2730

I've honestly tried ALOT of other ways including trying to calculate the acceleration of the first point on the graph and then substituting that for my a and that gave me weird answers also.

 

[collinsmark]

I assumed when the problem asked me for the x coordinate, it meant x as in position, but not x as in the x and y-axis on the graph below.

Yes, that's right. The x-component of the graph below is time. But the "x" that you are looking for is distance. So you are right.

From that, I set out to solve this problem using equation 2 and setting it up for final position like so:

x = v₀t + 1/2at² + x₀

Just make sure you are careful about how to interpret t in your equation. You are starting at t₀ = 30 sec, not zero seconds. So everything needs to be done with respect to t₀ = 30 sec as a starting point.

then since I knew I didn't have the acceleration for this problem, I tried to find it:

a = -40 - 20 / 50 - 30 = -3ft/s

'Looks good to me!

once I got the acceleration of -3ft/s I added my known values to the equation I setup:

x = 20*50 + (1/2*-3)50² + 20

Try it again, but this time be more careful about how you interpret t. The conditions x₀ = 20 ft, and v₀ = 20 ft/sec are true when t = 30 sec (not when t = 0 sec). You need to subtract off 30 sec from t to make sure everything lines up. In other words,

x = x₀ + v₀(t - 30) + (1/2)a(t - 30)²

 

[Meadman23]

Dude...WOW. I can't believe after countless hours all I needed to do was calculate the change in time. Thank you SO much!

As a clarification though, whenever I see t in a formula, does that mean I'm always supposed to find t - t₀? I never thought to do that because nothing I read up to this point seemed to suggest I should, that is unless I was trying to calculate the change in t as part of the formula to find average speed or average acceleration.

 

[collinsmark]

Dude...WOW. I can't believe after countless hours all I needed to do was calculate the change in time. Thank you SO much!

As a clarification though, whenever I see t in a formula, does that mean I'm always supposed to find t - t₀? I never thought to do that because nothing I read up to this point seemed to suggest I should, that is unless I was trying to calculate the change in t as part of the formula to find average speed or average acceleration.

Essentially, yes. One way to write that particular kinematics equation is

x = x₀ + v₀(Δt) + (1/2)a(Δt)²,

which is another way of saying,

x = x₀ + v₀(t - t₀) + (1/2)a(t - t₀)²

[Edit: Just make sure you understand what t₀ is. The constant t₀ is the specific time, within the interval of uniform acceleration, that that x = x₀ and v = v₀.]

 

[rwisz]

I would first clarify with the person asking the question what they mean by "x coordinate" - do they mean position or something else? If they do mean position, then the x coordinate at 50 seconds cannot be -180, as that would imply that the particle has moved backwards in space, which is not possible.

Assuming they do mean position, I would approach this problem by using the equations of motion you have listed. However, I would use the equation x = x0 + v0t + 1/2at^2, where x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration. This equation can be rearranged to solve for any of the variables, depending on what information you have.

In this case, we know the initial position (x0 = 20 ft), the initial velocity (v0 = 0 ft/s), and the time (t = 50 s). We do not know the acceleration, but we can calculate it using the equation a = (vf - v0)/t, where vf is the final velocity. Looking at the graph, we can see that the final velocity at 50 seconds is approximately 40 ft/s. Therefore, the acceleration is a = (40 - 0)/50 = 0.8 ft/s^2.

Plugging these values into the equation x = x0 + v0t + 1/2at^2, we get:

x = 20 + 0*50 + 1/2*0.8*50^2 = 20 + 0 + 1/2*2000 = 1020 ft

So the x coordinate of the particle at 50 seconds is 1020 feet. This makes more sense than -180 feet, as it shows that the particle has moved forward in space.

It's always important to check your answer and make sure it makes physical sense. In this case, the answer of 1020 feet is reasonable and consistent with the graph and the initial information given, whereas the answer of -180 feet does not make sense and is likely incorrect.