# Force and Constant Acceleration Problem

1. Feb 19, 2012

### sb13

1. The problem statement, all variables and given/known data
Sunjamming. A “sun yacht” is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of 780 kg and receives a push of 23 N. (a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

2. Relevant equations
F= ma
And the equations for motion with constant acceleration:
v=vo+at
x-xo= vot+ 1/2at^2
v^2= vo^2+2a(x-xo)
x-xo= 1/2(vo+v)t
x-xo=vt-1/2at^2

3. The attempt at a solution
I understand how to solve for the magnitude of acceleration in (a)
a= F/m = 23/780 =0.029487 m/s^2
But for the (b), I don't know how I would find the meters that the spacecraft travels in one day if all I have is acceleration which does have time units but they are squared, and then I would use one of the constant acceleration equations, but I need more information.
If I were to figure out the answer to (b) which is the time (t), then I think I could just plug it into one of those equations and easily get the velocity for (c).
Could someone help me with part (b) please?

Last edited: Feb 19, 2012
2. Feb 19, 2012

### LawrenceC

x-xo= vo+ 1/2at^2

This equation is incorrect. How can you add/subtract velocity from distance? Units of terms in equations MUST be uniform.

Use the correct version of this equation to determine how far it travels in a specified time.

3. Feb 19, 2012

### sb13

Those are the equations that my Physics book gave me. I did forget to include the t which is multiplied by the initial velocity, but I don't know where I could find the "correct" equations if those equations are not usable.

4. Feb 19, 2012

### LawrenceC

That is what I was getting at. You left the 't' out.

That equation enables one to compute the displacement when a object that may or may not be moving is accelerated at a constant rate. All you have to do is use it with the correct units.

5. Feb 19, 2012

### sb13

Do I need to convert seconds to days? Because I am not sure how to do that when the the units are s^2.

6. Feb 19, 2012

### LawrenceC

You have computed the acceleration in terms of m/sec^2. So if you choose to use those units in the equation

x = V0*t + 0.5*a*t^2

you must determine how many seconds there are in a day. If, on the other hand, you had determined the acceleration rate in terms of m/day^2, then you can plug it in directly. You will get the same answer.

The V0 accounts for the fact the object is already moving. If it is not, the term vanishes.

It cannot be overstressed how important units are when using equations. Many years ago when I was a student, I did not comprehend that. As I progressed in my studies and professional career, I realized how important units are. They keep you 'out of trouble'.

7. Feb 19, 2012

### sb13

Oh! Thank you so much. That was my main problem. I just was worried because I ended up with days^2 and I thought that would mess things up.

8. Feb 19, 2012

### sb13

So for (c) I tried the equation v=vo + at to find the velocity, but it is not giving the correct answer. The initial velocity is equal to zero, and the acceleration is equal to 0.029487 m/s^2, but I want the time to be in seconds so that it will cancel off one of the seconds leaving just m/s right? Which means the time equals one? That doesn't give the correct answer and that just means the acceleration is equal to the velocity. What do I do?

9. Feb 19, 2012

### LawrenceC

Did you determine how many seconds there are in a day?

10. Feb 19, 2012

### sb13

I got the correct distance of 1.1006 X 10^8, and for the time I converted seconds squared to days squared by multiplying by (3600^2)(24^2). In the equation of x-xo = vot+ 1/2at^2, I assumed that the initial velocity and initial distance was zero. Is that correct?

11. Feb 19, 2012

### LawrenceC

For the velocity with a zero initial velocity the equation reduces to

v = a*t

If a is .029 m/sec^2, the velocity is gotten from plugging the number of seconds in a day into the above equation.

12. Feb 19, 2012

### LawrenceC

I get something close to that but not that exactly. Your assumptions are correct.

Last edited: Feb 19, 2012
13. Feb 19, 2012

### sb13

You didn't get exactly 1.1006X10^8 for the distance?
But, I do need the velocity to be in m/s.

14. Feb 19, 2012

### LawrenceC

Here are my numbers:

x = .5 a t^2 = .5*.029*(3600*24)^2=1.08X10^8

Where do we differ?

15. Feb 19, 2012

### sb13

The only difference is that I carried out the significant digits on my acceleration.

(0.5)(0.029487)(3600*24)^2 = 1.1006 X 10^8

16. Feb 19, 2012

### LawrenceC

Good, then that part is finished. Did you get the velocity part yet?

17. Feb 19, 2012

### sb13

Well, I did end up getting the correct velocity. But I am wondering why I couldn't get the right answer with the v=vo + at formula

v=vo + at= (0.029487m/s^2)(1s)? Is one second the time? Then the velocity equals the acceleration, and that's not the right answer.

I tried v^2= vo^2 + 2a(x-xo), and that gave me the correct answer though. But, why do they give different answers? They aren't supposed to are they?

So v^2 = 0 + 2(0.029487 m/s^2)(1.1006X10^8 m) = 6.49068X10^6 m^2/s^2, so v = square root of 6.49068X10^6 = 2547.08 m/s

When I took the square root of the number, it took away the squares of the meters and seconds right?

18. Feb 19, 2012

### LawrenceC

The question asks how fast will it be moving after a day of accelerating.

19. Feb 19, 2012

### sb13

Ok. So does that mean that the equation that I choose to use to determine the velocity should include the distance (x)?

20. Feb 19, 2012

### LawrenceC

Yes. If you use the other equation where V = at you'll get the same result.

V = .029487*(3600*24)=2547.6 m/sec

V = sqrt(2*.029487*1.1006X10^8)=2547.6 m/sec