Calculating Particle Motion with Varying Force: Solving for Distance and Time

[LukeEvans]

A particle of mass m travels in a straight line and experiences a single force, F(t) , in the direction of motion that varies with time t.
( a ) Given that m = 6 kg and F( t ) = 6 sin t + 24 N, use Newton’s 2nd law to show that the acceleration, a m/s2 , of the particle is,
a = sin t + 4
( 2 marks )
( b ) Given that initially the particle starts from rest at a particular datum point on the line of motion, show that the distance traveled , x , from the datum point by the particle at time t is,
x = 2 t2 + t - sin t
Hence, show that the time taken for the particle to travel 8 m from the datum point is given by the solution of the equation f( t ) = 0 where
f( t ) = 2 t2 + t - 8 - sin t
( 4 marks )


The above is the question I'm struggling with. I have completed the first part (worth two marks) but included it for reference.

The result I got from the first part was (6 sin t + 24N)/6 = sin t + 4 m/s2.

The second part (worth four marks) has got me stumped. I would dearly love to see the steps taken to satisfy the question. Thank you.

 

[gneill]

How familiar are you with calculus? If you integrate acceleration over time, what do you get?

 

[LukeEvans]

How familiar are you with calculus?

Reasonably familiar with differentiations/integrations i.e. maths for graphics.

If you integrate acceleration over time, what do you get?

A change in acceleration would be my best guess.

 

[gneill]

Integrate, not differentiate. When you integrate acceleration you get velocity. Integrate velocity and you get distance.

 

[LukeEvans]

Excellent. Would you be able to show this in action in terms of part (b) above?

 

[gneill]

Excellent. Would you be able to show this in action in terms of part (b) above?

Forum policy dictates that you should make an attempt before help is given, and then as guidance towards a solution.

How about you start by attempting to find the expression for the velocity? You have already obtained the acceleration,

a(t) = sin(t) + 4

Can you integrate this function with respect to time?

 

[LukeEvans]

OK so to integrate a(t) = sin(t) + 4

I would write a(t) = -cos(t) + 4

Am I wrong?

 

[gneill]

OK so to integrate a(t) = sin(t) + 4

I would write a(t) = -cos(t) + 4

Am I wrong?

Well, let's just say that you're not entirely right.

The new function that is obtained by integrating a(t) is no longer "a(t)", it's something new. Let's call it v(t)

You pegged the integral of the sin() function, but missed out on integrating constant (4). The integral of a constant yields t multiplied by that constant, and while we're at it, you need to add a new, undetermined constant, say C, if you're doing "indefinite" integrals.

So $\int sin(t) + 4 \; dt \rightarrow -cos(t) + 4t + C$

You can dispense with the constant C if you turn this into a definite integral (with set limits of integration). You're interested in knowing the velocity at time t starting from time t = 0, so to sum up,

$$v(t) = \int_0^t a(t) \; dt = \int_0^t sin(t) + 4 \; dt = -cos(t) + 4t + 1$$
$$v(t) = -cos(t) + 4t + 1$$

Can you now take v(t) and integrate it in a similar fashion?

 

[LukeEvans]

Excellent.

I'd say then that the integral of that is

d(t) = -sin(t) + 4t + C

Any good?

 

[gneill]

Excellent.

I'd say then that the integral of that is

d(t) = -sin(t) + 4t + C

Any good?

Try again. Each term in the expression must be integrated. The terms are:

-cos(t) --> -sin(t) {okay, you got this one!}

4t --> ?

1 --> ?

 

[LukeEvans]

Try again. Each term in the expression must be integrated. The terms are:

-cos(t) --> -sin(t) {okay, you got this one!}

4t --> ?

1 --> ?

Is it then

-sin(t) + 2t2 + c

 

[gneill]

Is it then

-sin(t) + 2t2 + c

Very close. It's

x(t) = -sin(t) + 2t² + t + c

You had forgotten to integrate the constant 1. It becomes t; integral of 1 --> t.

Now, if you look at the original problem statement you will see that they mention that the particle starts from some "particular datum point". Which term in your position function x(t) would you say corresponds to that initial position?

 

[LukeEvans]

x(t) = -sin(t) + 2t2 + t + c

is it the t term? as in t could be 0, 1, and so on?

 

[gneill]

t is the time variable. The initial time is t = 0. The initial position occurs when t = 0. What's x(0)?

 

[LukeEvans]

t is the time variable. The initial time is t = 0. The initial position occurs when t = 0. What's x(0)?

x(0) would be distance zero? As in the datum point 0?

 

[gneill]

Yes, the distance would be zero. But you have derived x(t), which is the position at time t. The distance traveled from some initial position would be x(t) - x0, where x0 is the "initial datum" or reference point.

If you plug t=0 into the equation you found for x(t), what is the result?

 

[LukeEvans]

Yes, the distance would be zero. But you have derived x(t), which is the position at time t. The distance traveled from some initial position would be x(t) - x0, where x0 is the "initial datum" or reference point.

If you plug t=0 into the equation you found for x(t), what is the result?

x(0) = -sin(0) + 0 + 0 + c

or just x(0) = 0 + c

Correct?

 

[gneill]

Right. x(0) = c, c being the integration constant that you don't have a value for (until now!). So c is the initial position. Call it x0. Your equation for position is thus:

x(t) = x0 - sin(t) + 2t²+ t

Now, at some time t the position is x(t). The initial position is x0. So the distance of the particle from x0 at time x(t) is d(t) = x(t) - x0

If you move the x0 from the right hand side to the left had side of your position equation, you'll have the expression for the distance at time t. You should then be in a position to complete the problem.

 

[LukeEvans]

Right. x(0) = c, c being the integration constant that you don't have a value for (until now!). So c is the initial position. Call it x0. Your equation for position is thus:

x(t) = x0 - sin(t) + 2t²+ t

Now, at some time t the position is x(t). The initial position is x0. So the distance of the particle from x0 at time x(t) is d(t) = x(t) - x0

If you move the x0 from the right hand side to the left had side of your position equation, you'll have the expression for the distance at time t. You should then be in a position to complete the problem.

x(t)+x0 = 2t²+ t - sin(t)

is this what you mean? I've got to show that f( t ) = 2 t2 + t - 8 - sin t

 

[LukeEvans]

I can see it's almost looking like it...

 

[gneill]

x(t)+x0 = 2t²+ t - sin(t)

is this what you mean? I've got to show that f( t ) = 2 t2 + t - 8 - sin t

Reread the last paragraph of the problem statement. We've just finished deriving the function that describes the distance for the particle with respect to time,
$$d(t) = 2t^2 + t - sin(t)$$
What information does the problem statement supply?

 

[LukeEvans]

Is the next step to include the 8m (which is the information loaded in the last paragraph?)?

So it should be:

d(t) = 2t² + t -8− sin(t)

 

[LukeEvans]

Actually as a side note, I'm mulling a question over. Would the derivative of the position (distance) give the velocity?

As giving the anti derivative of velocity gives us position, would the derivative of position give us the velocity?

 

[gneill]

Is the next step to include the 8m (which is the information loaded in the last paragraph?)?

So it should be:

d(t) = 2t² + t -8− sin(t)

d *IS* the distance. That was the whole point of deriving an equation for distance.

So at some as yet unknown time t, d(t) will be 8m. That is,

8m = 2t² + t − sin(t)

Which you can then rearrange to put in the form of a function whose root ( or possibly roots ) will give you the time t: f(t) = 2t² + t - 8 − sin(t).

There are good reasons for doing it this way, particularly if the equation happens to be one whose roots cannot be found by algebraic manipulation alone.

 

[LukeEvans]

Ah I believe I understand now. Thank you for helping me understand this, or at least figure out what to do in an exam!

I have one more issue - should I be asked to integrate 4e^2t + 2 how would I deal with the e?

My best guess would be that the integral of that would be:

2t + e^2t + C

 

[gneill]

Ah I believe I understand now. Thank you for helping me understand this, or at least figure out what to do in an exam!

I have one more issue - should I be asked to integrate 4e^2t + 2 how would I deal with the e?

My best guess would be that the integral of that would be:

2t + e^2t + C

Close.

2t + 2e^2t + C

You can verify your integration attempts online using the Wolfram online integration app. Do a web search for Wolfram Integrator.

 

[LukeEvans]

Wow thank you. Glad I was close, and thanks for the heads up too!