# Acceleration of a sled

1. Jan 24, 2010

### mybrohshi5

1. The problem statement, all variables and given/known data

A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time the sled passes a red flag stuck into the snow at an unknown distance from the top of the hill; 2.00 s after passing the red flag the sled is 17.83 m past the red flag; and another 2.00 s later, the sled is now 42.27 m past the red flag.

What is the acceleration of the sled?
What is the speed of the sled when it passes the red flag?
How much time did it take for the sled to go from the top to the red flag?
How far did the sled go during the 1st second after passing the red flag?
How far did the sled go during the 2nd second after passing the red flag?What is the distance from the top of the hill to the red flag?

2. Relevant equations

constant acceleration --> x(t) = x + (v)(t) + 1/2(a)(t^2)
d = vi(t) + 1/2(a)(t^2)

3. The attempt at a solution

First i need to find the acceleration of the sled.

Would i set it up as 47.27-17.83 = (0)(2) + 1/2(a)2^2
so a = 14.72 m/s ???? is that correct?

any help would be greatly appreciated. thank you

Last edited: Jan 24, 2010
2. Jan 24, 2010

### PhanthomJay

I'm not sure where your numbers are coming from or what formula you are using, but your answer is incorrect. You need to use the kinematic motion equations for constant acceleration. Please list the relevant equation(s). You are given a couple of sets of data values for displacement and time.

3. Jan 24, 2010

### mybrohshi5

Relevant equations and attempt at a solution edited.

can anyone confirm that 14.72 m/s is the right acceleration?

Thank you

4. Jan 25, 2010

### mybrohshi5

Anyone? this hw problem is due tomorrow and i would really like to figure it out =)

thanks!

5. Jan 25, 2010

### mybrohshi5

14.72 is wrong and i have no idea how to find the correct acceleration with the given information.

Can anyone lead me in the right direction please!

6. Jan 25, 2010

### cdotter

I'm learning kinematics as well, but I think you need to look at the entire distance and time traveled.

$(42.27m)=\frac{1}{2}a(2.00s + 2.00s)^2$

Last edited: Jan 25, 2010
7. Jan 25, 2010

### mybrohshi5

thank you for the help but thats incorrect :(

i solved for a=5.28 m/s^2 and it says its wrong.

i dont think that works because the time from the start to the first red flag is unknown......

anybody else? i really need to figure this one out. its my last one and its really bothering me haha!!!!

thank you

8. Jan 25, 2010

### PhanthomJay

You have the correct relevant equation; apply it first at d=17.03 when t = ______and then apply it again at d =42.27 when t = _______. (Fill in the blanks). Then you have 2 equations with 2 unknowns, which you can solve.

9. Jan 25, 2010

### cdotter

Why doesn't my way work? Your way is finding the average acceleration, right?

10. Jan 25, 2010

### mybrohshi5

would the time for both be 2 seconds? or would the first one be 2 and the second one be 4?

thank you

11. Jan 25, 2010

### PhanthomJay

No.
yes.

Don't forget that the initial speed at the red flag is unknown. Both the equations are referenced to the red flag point.

12. Jan 25, 2010

### PhanthomJay

No, I'm finding the instantaneous constant acceleration, which is also the average if it's constant. Your way assumes that the initial speed at the red flag is 0. That is not correct; it is moving at that point.

Last edited: Jan 25, 2010
13. Jan 25, 2010

### mybrohshi5

ok thank you.

can you confirm if this is correct really fast?

17.83=vi(4) + .5a(2^2)
solved for a
a= (17.83 - vi(2)) / 2

plugged this a into the other equation

42.27=vi(4) + .5((17.83 - 2vi) / 2)(4^2)

i solved for vi=7.2625 m/s

and then plugged that into one of the two equations and got

a= 1.6525 m/s^2 for my acceleration but this seemed really slow to me?

is the acceleration correct?

Last edited: Jan 25, 2010
14. Jan 25, 2010

### PhanthomJay

correct the typo, it's vi(2), which you have calculated correctly below
looks good. It's tough to judge speed from an acceleration value. Actually its not going all that slow, nor is it's acceleration low. A commercial jet aircraft takes off with an acceleration of about 3m/s^2 or less.

15. Jan 25, 2010

### mybrohshi5

oh wow thats good to know. it will be good to keep in mind when doing these problems.

thank you for all of your help. i really learned a lot :)

16. Jan 25, 2010

### mybrohshi5

Now im stuck comprehending the second part of the question.

What is the speed of the sled when it passes the red flag?

speed is distance/time

how can speed be found if the distance and time from the starting point at rest to the red flag is unknown?

17. Jan 25, 2010

### mybrohshi5

Would the answer maybe be 12.215 m/s?

i got this from finding v=17.83/2 = 8.915 m/s

then vf = 8.915 + 1.65(2) = 12.215 m/s

18. Jan 25, 2010

### mybrohshi5

ok well its not 12.215 m/s like i thought :(

19. Jan 25, 2010

### PhanthomJay

I have to sign off for now. But here's a hint: you already calculated the speed of the sled at the flag , in your post #13 above!

20. Jan 25, 2010

### mybrohshi5

thank you. i didnt notice that cause i thought speed and velocity initial were different but i guess in this case they are the same.

thank you very much for all your time and help.

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