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Physics cp2.63 find the initial velocity

karush

Well-known member
Jan 31, 2012
2,725
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...
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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:
\(\displaystyle v(t) - v(0) = \int_0^t a(t') ~ dt'\)

\(\displaystyle x(t) - x(0) = \int_0^t v(t') ~ dt'\)

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

-Dan
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.
 

karush

Well-known member
Jan 31, 2012
2,725
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 $
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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...
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
 

karush

Well-known member
Jan 31, 2012
2,725
ok lets drop this post there might be a typo in this..
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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.
 

karush

Well-known member
Jan 31, 2012
2,725
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?

 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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:
$$a(t)=-2.00 \, m/s^2 +3.00 \, m/s^3\cdot t$$
 

karush

Well-known member
Jan 31, 2012
2,725
ok lets 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
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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