Motion of a Particle

[Jason76]

Starting at \(\displaystyle t = 0\) a particle moves along the \(\displaystyle x\) axis so that it's position at time \(\displaystyle t\) is given by \(\displaystyle x(t) = t^{4} - 5t^{2} + 2t\). For what values of \(\displaystyle t\) does that particle take while moving to the left?

Answer: \(\displaystyle 0.203 < t < 1.470\)

Any starting hints on how this came to be?

I know that the derivative of position is velocity. Perhaps this relates somehow.

 

[JeffM]

Starting at \(\displaystyle t = 0\) a particle moves along the \(\displaystyle x\) axis so that it's position at time \(\displaystyle t\) is given by \(\displaystyle x(t) = t^{4} - 5t^{2} + 2t\). For what values of \(\displaystyle t\) does that particle take while moving to the left?

Answer: \(\displaystyle 0.203 < t < 1.470\)

Any starting hints on how this came to be?

I know that the derivative of position is velocity. Perhaps this relates somehow. It does

I doubt you have given the complete problem. Nor have you explained the topic currently being studied.

If the sign of the instantaneous velocity is positive, what direction is the particle moving at that instant, right or left?

If the sign of the instantaneous velocity is negative, what direction is the particle moving at that instant?

If the instantaneous velocity is zero, what direction is the particle moving at that instant?

At t = 0, what is the sign of the instantaneous velocity.

At t = 1, what is the sign of the instantaneous velocity.

At t = 2, what is the sign of the instantaneous velocity.

 

[Jason76]

I doubt you have given the complete problem. Nor have you explained the topic currently being studied.

If the sign of the instantaneous velocity is positive, what direction is the particle moving at that instant, right or left?

If the sign of the instantaneous velocity is negative, what direction is the particle moving at that instant?

If the instantaneous velocity is zero, what direction is the particle moving at that instant?

At t = 0, what is the sign of the instantaneous velocity.

At t = 1, what is the sign of the instantaneous velocity.

At t = 2, what is the sign of the instantaneous velocity.

I have the complete problem there.

\(\displaystyle x(t) = (0)^{4} - 5(0)^{2} + 2(0) = 0\)

\(\displaystyle x(t) = (1)^{4} - 5(1)^{2} + 2(1) = -2\) - negative

\(\displaystyle x(t) = (2)^{4} - 5(2)^{2} + 2(2) = -8\) - negative

 

[HallsofIvy]

JeffM asked you six questions, designed to help you think about this question. The first three, what does sign of the velocity mean, you did not answer. The last three, what is the velocity at x= 0, 1, and 2, you answered incorrectly because you used the position function, not the velocity function. Since the position function is a fourth degree polynomial, it derivative, the velocity function, will be a third degree polynomial. You may find it difficult to determine where it is negative. That is why JeffM asked if you were sure you had stated the problem competely.

 

[Jason76]

Using a "conventional" coordinate system, "moving to the left" would mean "velocity is negative". The velocity function is the derivative of the position function. So this problem appears to be asking you to differentiate \(\displaystyle t^4- 5t^2+ 2t\) and then find the values of t for which that derivative is negative.

So, to start what is the derivative of \(\displaystyle t^4- 5t^2+ 2t\)?

\(\displaystyle s(t) = t^4- 5t^2+ 2t\) - position

\(\displaystyle \dfrac{ds}{dt} = v(t) = 4t - 10t + 2\) - velocity

Ok, which test values do you put in?

 

[DrPhil]

\(\displaystyle s(t) = t^4- 5t^2+ 2t\) = position

\(\displaystyle \dfrac{ds}{dt} = v(t) = 4t^3 - 10t + 2\) = velocity

Ok, which test values do you put in?

I think you had a typo - left off power "t^3" in 1st term of v.

Do you affirm that "moving left" is negative velocity? If so, solve the inequality

\(\displaystyle 4t^3 - 10t + 2 < 0\)

for t. Since the expression is cubic, you may have to hunt for the roots by successive approximations or by Newton's method. They seem to be assuming that only positive \(\displaystyle t\) is relevant. Draw a graph of v(t) to get started in the right ballpark.

 

[JeffM]

I have the complete problem there.

What made me doubt that we have the complete problem is that the answers you say are given in the book are not exact answers. Maybe there was something in the statement of the problem about being correct to the nearest thousandth? Furthermore, finding an approximate solution to this problem may involve the Newton-Raphson method, which could be indicated in the question itself or by the context of the topic that you are currently studying.

I sometimes think that we put a great deal more care into writing our responses to you than you do into trying to comprehend them.

 

[Jason76]

I think you had a typo - left off power "t^3" in 1st term of v.

Do you affirm that "moving left" is negative velocity? If so, solve the inequality

\(\displaystyle 4t^3 - 10t + 2 < 0\)

for t. Since the expression is cubic, you may have to hunt for the roots by successive approximations or by Newton's method. They seem to be assuming that only positive \(\displaystyle t\) is relevant. Draw a graph of v(t) to get started in the right ballpark.

a). For what values does \(\displaystyle t\) take on while moving to the left?

No \(\displaystyle x(t) = t^{4} - 5t^{2} + 2t\) is the equation.

\(\displaystyle s(t) = t^{4} - 5t^{2} + 2t\) - position function

\(\displaystyle \dfrac{ds}{dt} = v(t) = t^{3} - 10t + 2\)

\(\displaystyle v(t) = t^{3} - 10t + 2 < 0\) - Since negative values are what we are seeking.

So your saying use successive approximations or Newton's method. But with what values do we do the successive approximations?

Also, if the t is moving to the left, then perhaps we need to know some starting point. What exactly are the independent and dependent variables regarding particle motion? I know that when we are discussing \(\displaystyle t\) and a baseball being hit, then \(\displaystyle t\) is time, and \(\displaystyle s(t)\) is the distance (in regards to the position function).

Summary of Baseball functions

\(\displaystyle t\) is time, and \(\displaystyle s(t)\) is the distance (in regards to the position function).

\(\displaystyle t\) is time, and \(\displaystyle v(t)\) is the velocity (in regards to the velocity function).

\(\displaystyle t\) is time, and \(\displaystyle a(t)\) is the acceleration (in regards to the acceleration function).

 

[DrPhil]

a). For what values does \(\displaystyle t\) take on while the particle is moving to the left?

No \(\displaystyle x(t) = t^{4} - 5t^{2} + 2t\) is the equation.

\(\displaystyle s(t) = t^{4} - 5t^{2} + 2t\) - position function These two are identical, unless
....................................you are considering s to be a one-dimensional vector
....................................and x to be the one and only component of that vector

\(\displaystyle \dfrac{ds}{dt} = v(t) = t^{3} - 10t + 2\)

\(\displaystyle v(t) = t^{3} - 10t + 2 < 0\) - Since negative velocities are what we are seeking.

So your saying use successive approximations or Newton's method. But with what values do we do the successive approximations?
When in doubt, draw a graph to find a reasonable starting point for the root calculation.
A set of times for drawing the graph could be {0, .5, 1, 1.5, 2} or as many more as you want.

Also, if the particle is moving to the left, then perhaps we need to know some starting point. What exactly are the independent and dependent variables regarding particle motion? I know that when we are discussing \(\displaystyle t\) and a baseball being hit, then \(\displaystyle t\) is time, and \(\displaystyle s(t)\) is the distance (in regards to the position function).
s(0) = 0, so that would appear to be the starting point. In the answer, they ignored negative t.

Summary of Baseball functions Note that these are all vectors

\(\displaystyle t\) is time, and \(\displaystyle \ \vec{s(t)}\) is the distance (in regards to the position function).

\(\displaystyle t\) is time, and \(\displaystyle \ \vec{v(t)}\) is the velocity (in regards to the velocity function).

\(\displaystyle t\) is time, and \(\displaystyle \ \vec{a(t)}\) is the acceleration (in regards to the acceleration function).

.

 

[Jason76]

Maybe we are thinking to deep on this one. Perhaps the answer is to simply graph the function :-? Then you can see what values the function takes as it goes left from \(\displaystyle 0\).

But the problem is that a graph goes on forever to the left. So how can there be a left interval?

But perhaps this is some kind of flipped graph, where time is on the \(\displaystyle y\) axis rather than the \(\displaystyle x\) one.

Yeah, I think that's the problem.

 

[Bob Brown MSEE]

Using graphs for insight

Maybe we are thinking to deep on this one. Perhaps the answer is to simply graph the function :-?

"thinking to deep" Yes!!!!!!!!!!!!!!
Look first at graph for insight.
Click Here

"Graph is flipped" Yes!!!!!!!!!!!!!!!
Great insight. It is kind of strange that the problem refers to "moving to the left" when speaking about "x(t)" other than the common x-axis experience. Here the problem makes x(t) the dependent variable (on y-axis), so "left" should be interpreted as "DOWN"

Another insight is the answer given is at the local max and min of x(t). Click Here and Here. That is the only region for t>0 when x(t) is going "DOWN".

Wonderful example of using a graphing calculator to get in-sight... good work!


Post Script: Doing the problem to hand-in.
Simply use the derivative to find the local max and min as you were taught in class.

 

[Deleted member 4993]

Maybe we are thinking to deep on this one. Perhaps the answer is to simply graph the function :-? Then you can see what values the function takes as it goes left from \(\displaystyle 0\).

But the problem is that a graph goes on forever to the left. So how can there be a left interval?


No it does not - it stops at t = 0. Particle motion is not defined at t<0.


But perhaps this is some kind of flipped graph, where time is on the \(\displaystyle y\) axis rather than the \(\displaystyle x\) one.

Yeah, I think that's the problem.

.

 

[Jason76]

This one is still tough, though the other problems (I posted) are making more sense. I guess I will dwell over it some more.