Exercise: position vector of a particle

[Kernul]

Homework Statement

The acceleration of a particle moving only on a horizontal xy plane is given by a = 3ti + 4tj, where a is in meters per second squared and t is in seconds. At t = 0, the position vector r = (20.0 m)i + (40.0 m)j locates the particle, which then has the velocity vector v = (5.00m/s)i + (2.00m/s)j. At t = 4.00s, what are
(a) its position vector in unit-vector notation and
(b) the angle between its direction of travel and the positive direction of the x axis?

Homework Equations

x = x₀ + v_ox * t + 1/2 * a_x * t²
y = y₀ + v_oy * t + 1/2 * a_y * t²

The Attempt at a Solution

So, since a_x = 3t and a_y = 4t, I got the following values for the position vector:
(136m)i + (176m)j

The problem is that the values aren't the solution given by the book (which is (72m)i + (90.7m)j ).
I keep reading but nothing comes to my mind besides the fact that probably I did something wrong with the acceleration.

 

[SteamKing]

Homework Statement

The acceleration of a particle moving only on a horizontal xy plane is given by a = 3ti + 4tj, where a is in meters per second squared and t is in seconds. At t = 0, the position vector r = (20.0 m)i + (40.0 m)j locates the particle, which then has the velocity vector v = (5.00m/s)i + (2.00m/s)j. At t = 4.00s, what are
(a) its position vector in unit-vector notation and
(b) the angle between its direction of travel and the positive direction of the x axis?

Homework Equations

x = x₀ + v_ox * t + 1/2 * a_x * t²
y = y₀ + v_oy * t + 1/2 * a_y * t²

The Attempt at a Solution

So, since a_x = 3t and a_y = 4t, I got the following values for the position vector:
(136m)i + (176m)j

The problem is that the values aren't the solution given by the book (which is (72m)i + (90.7m)j ).
I keep reading but nothing comes to my mind besides the fact that probably I did something wrong with the acceleration.

The equations you cited in Section 2 are valid only for constant acceleration. Your acceleration is changing with time.

 

[Kernul]

The book doesn't give any example with a non-constant acceleration. I don't know how to proceed in this case.
Could you guide me to the solution?

 

[SteamKing]

The book doesn't give any example with a non-constant acceleration. I don't know how to proceed in this case.
Could you guide me to the solution?

Are you sure?

Have you seen these expressions:

v = ds / dt

a = dv /dt

 

[Kernul]

Yes, I know them.
The last one is an instantaneous acceleration, right?

 

[SteamKing]

Yes, I know them.
The last one is an instantaneous acceleration, right?

Yes, it is.

When acceleration is changing with time, you must integrate acceleration w.r.t. time to find the instantaneous velocity, and then integrate velocity w.r.t. time to find position.

v(t) = ∫ a(t) dt + v₀

s(t) = ∫ v(t) dt + v₀ ⋅ (t) + s₀

where v₀ and s₀ represent the initial velocity and position, respectively.

 

[Kernul]

If from the second equation I simply write
s(t) = ∫ v(t) dt + s0 I have the solution.
Is it okay? Because if I do the second equation as you wrote it, it has a +20 on the first component and a +8 on the second component. I guess it was a typo from you.

 

[SteamKing]

If from the second equation I simply write
s(t) = ∫ v(t) dt + s0 I have the solution.
Is it okay? Because if I do the second equation as you wrote it, it has a +20 on the first component and a +8 on the second component. I guess it was a typo from you.

s₀ can be written in terms of a position vector with i and j components.

All of the integrations can be performed separately to obtain the i and j components of the velocity and the position of the particle as a function of time.

 

[Kernul]

Yeah, I know. The last equation you wrote would have these numbers:

s(t) = (3 * t³) / 6 i + (4 * t³) / 6 j + 5 * t i + 2 * t j + 5 * t i + 2 * t j + 20 i + 40 j

Am I right?

 

[SteamKing]

Yeah, I know. The last equation you wrote would have these numbers:

s(t) = (3 * t³) / 6 i + (4 * t³) / 6 j + 5 * t i + 2 * t j + 5 * t i + 2 * t j + 20 i + 40 j

Am I right?

I think you have included v₀ ⋅ t twice in the expression for s(t). You'll need to evaluate s(t) for t = 4 sec.

 

[Kernul]

That's exactly what I wanted to make you notice. In your equations I have two of them:

v(t) = ∫ a(t) dt + v₀

s(t) = ∫ v(t) dt + v₀ ⋅ (t) + s₀

With:
v(t) = (3 * t²) / 2 i + (4 * t²) / 2 j + 5 i + 2 j
and
s(t) = (3 * t³) / 6 i + (4 * t³) / 6 j + 5 * t i + 2 * t j + (5 i + 2 j) * t + 20 i + 40 j

 

[SteamKing]

That's exactly what I wanted to make you notice. In your equations I have two of them:

With:
v(t) = (3 * t²) / 2 i + (4 * t²) / 2 j + 5 i + 2 j
and
s(t) = (3 * t³) / 6 i + (4 * t³) / 6 j + 5 * t i + 2 * t j + (5 i + 2 j) * t + 20 i + 40 j

v(t) = ∫ a(t) dt + v₀

∫ v(t) dt = s(t)

∫ v(t) dt = ∫ [∫ a(t) dt + v₀] dt = ∫∫ a(t) dt dt + ∫ v₀ dt = s(t) + v₀ ⋅ t + s₀

Since:
v₀ = 5 i + 2 j

If you multiply v₀ by t, then you have 5 ⋅ t i + 2 ⋅ t i

You are doing only one integration where v₀ is present, not two.

 

[Kernul]

Wait, I'm getting confused.
First we do this integral:
v(t) = ∫ a(t) dt + v₀ = (3 * t²) / 2 i + (4 * t²) / 2 j + 5 i + 2 j
after this, we do this other integral:
s(t) = ∫ v(t) dt + v₀ ⋅ (t) + s₀ = ∫ (3 * t²) / 2 i + (4 * t²) / 2 j + 5 i + 2 j dt + v₀ ⋅ (t) + s₀
Am I right? Because this is what comes out from your equations.

 

[SteamKing]

Wait, I'm getting confused.
First we do this integral:
v(t) = ∫ a(t) dt + v₀ = (3 * t²) / 2 i + (4 * t²) / 2 j + 5 i + 2 j
after this, we do this other integral:
s(t) = ∫ v(t) dt + v₀ ⋅ (t) + s₀ = ∫ (3 * t²) / 2 i + (4 * t²) / 2 j + 5 i + 2 j dt + v₀ ⋅ (t) + s₀
Am I right? Because this is what comes out from your equations.

No, it's not.

v₀ is a constant vector, namely v₀ = 5i + 2j. It does not depend on t.

When integrating a(t), v₀ is added after the integration is done, which is why v(t) = ∫ a(t) dt + v₀

When integrating v(t), then another constant of integration, which is called s₀, is added after the integration of v(t) is done.
But because v(t) includes v₀, when v(t) is integrated, the v₀ term must be integrated at well. Since v₀ is a constant, its integral is just v₀ ⋅ t.

s(t) = ∫ (v(t) + v₀) dt = ∫ v(t) dt + ∫ v₀ dt = ∫ v(t) dt + v₀ ⋅ t

I think your confusion stems from the fact that the v(t) inside the integral sign involves only the part of v(t) which varies with time.

I think these equations will be more clear:

v(t) = ∫ a(t) dt = v₁(t) + v₀

s(t) = ∫ v (t) dt = ∫ (v₁(t) + v₀) dt = ∫ v₁(t) dt + ∫ v₀ dt = ∫ v₁(t) dt + v₀ ⋅ t + s₀

 

[Kernul]

Oooh! So the second equation was simply "s(t) = ∫ v (t) dt".
Okay, I got it now.
Thank you!