MHB Cp2.63 find the initial velocity

AI Thread Summary
The discussion revolves around finding the initial velocity of a particle given its acceleration function, which is questioned for its correctness due to unit inconsistencies. Participants clarify that the acceleration should not combine terms with different units, suggesting a possible typo in the original equation. They emphasize the need to integrate acceleration to find velocity and then position, ultimately setting conditions for the particle's position at two different times. The conversation highlights confusion over the acceleration's formulation and the importance of correctly interpreting physical equations. The participants agree that resolving the unit issue is crucial for accurately solving the problem.
karush
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cp2.63. The acceleration of a particle is given by
$$a(t)=-2.00 \, m/s^2 +3.00 \, m/s^3.$$
a. find the initial velocity $v_0$ such that the particle will have the same x-coordinate at $t=4.00\, s$ as it had at $t=0$.
b. What will be the velocity at $t=4.00 s$

ok sorry but I don't even know step one...
 
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karush said:
cp2.63. The acceleration of a particle is given by
$$a(t)=-2.00 \, m/s^2 +3.00 \, m/s^3.$$
a. find the initial velocity $v_0$ such that the particle will have the same x-coordinate at $t=4.00\, s$ as it had at $t=0$.
b. What will be the velocity at $t=4.00 s$

ok sorry but I don't even know step one...
Well, start from the beginning:
[math]v(t) - v(0) = \int_0^t a(t') ~ dt'[/math]

[math]x(t) - x(0) = \int_0^t v(t') ~ dt'[/math]

Once you get that then go ahead and set x(0) = x(4) and see what you need to have for v(0).

-Dan
 
I would think that "step one" would be to recognize that "-2.0 m/s2+ 3.0 m/s2" is just 1.0 m/s2! Then use, as Topsquark suggests, that velocity is the "anti-derivative" of acceleration and distance moved is the "anti-derivative" of velocity.
 
HallsofIvy said:
I would think that "step one" would be to recognize that "-2.0 m/s2+ 3.0 m/s2" is just 1.0 m/s2! Then use, as Topsquark suggests, that velocity is the "anti-derivative" of acceleration and distance moved is the "anti-derivative" of velocity.
so would it be this?
$\displaystyle \int_0^t v(t') ~ dt' =\int_0^4 t\, dt $
 
karush said:
cp2.63. The acceleration of a particle is given by
$$a(t)=-2.00 \, m/s^2 +3.00 \, m/s^3.$$
a. find the initial velocity $v_0$ such that the particle will have the same x-coordinate at $t=4.00\, s$ as it had at $t=0$.
b. What will be the velocity at $t=4.00 s$

ok sorry but I don't even know step one...

HallsofIvy said:
I would think that "step one" would be to recognize that "-2.0 m/s2+ 3.0 m/s2" is just 1.0 m/s2! Then use, as Topsquark suggests, that velocity is the "anti-derivative" of acceleration and distance moved is the "anti-derivative" of velocity.
Okay, I have to ask this, then, before we go further. Is the acceleration -2.00 m/s^2 + 3.00 m/s^3 or -2.00 m/s^2 + 3.00 m/s^2 ?

-Dan
 
ok let's drop this post there might be a typo in this..
 
I think the velocity is *actually* non-linear. I expect that there is a jerk in the progression of cause to effect. Likely the time is missing in the second term. With a time factor the particle will wibble and wobble.
The unit of the second term is $\text{m}/\text{s}^3$ after all.
 
Klaas van Aarsen said:
I think the velocity is *actually* non-linear. I expect that there is a jerk in the progression of cause to effect. Likely the time is missing in the second term. With a time factor the particle will wibble and wobble.
The unit of the second term is $\text{m}/\text{s}^3$ after all.

yes the OP equation is correct I looked up in the book

however not real sure how a plot of this would look like would
$y=-2x^2+3x^3$ on desmos give us the graph?

[DESMOS]advanced: {"version":7,"graph":{"showGrid":false,"xAxisStep":1,"squareAxes":false,"viewport":{"xmin":-1.481,"ymin":-1.718,"xmax":12,"ymax":1000}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"y=-2x^2+3x^3\\ \\left\\{0\\le x\\le6\\right\\}"}]}}[/DESMOS]
 
Last edited:
karush said:
yes the OP equation is correct I looked up in the book

It cannot be correct.
It says:
$$a(t)=-2.00 \, \text{m}/\text{s}^2 +3.00 \, \text{m}/\text{s}^3$$
But we cannot add $\text{m}/\text{s}^2$ to $\text{m}/\text{s}^3$.
It does not make sense to split the terms like that either.

Likely it should be:
$$a(t)=-2.00 \, m/s^2 +3.00 \, m/s^3\cdot t$$
 
  • #10
ok let's work with that
its basically a learning problem anyway

so is the reason you do not have t in the first term is bc that is the acceleration at t=0
 
  • #11
Klaas van Aarsen said:
It cannot be correct.
It says:
$$a(t)=-2.00 \, \text{m}/\text{s}^2 +3.00 \, \text{m}/\text{s}^3$$
But we cannot add $\text{m}/\text{s}^2$ to $\text{m}/\text{s}^3$.
It does not make sense to split the terms like that either.

Likely it should be:
$$a(t)=-2.00 \, m/s^2 +3.00 \, m/s^3\cdot t$$
Unless the "3" carries units of distance. I've seen this "unit drop" often when moving Physics concepts to Mathematics.

-Dan
 
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