# Acceleration of a sled

## Homework Statement

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?

## Homework Equations

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

## 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

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## Answers and Replies

PhanthomJay
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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.

Relevant equations and attempt at a solution edited.

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

Thank you

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

thanks!

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!

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

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

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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

PhanthomJay
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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
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.

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.

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

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.

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

thank you

PhanthomJay
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would the time for both be 2 seconds?
No.
or would the first one be 2 and the second one be 4?
yes.

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

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

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

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?

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PhanthomJay
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ok thank you.

can you confirm if this is correct really fast?

17.83=vi(4) + .5a(2^2)
correct the typo, it's vi(2), which you have calculated correctly below
solved for a= (17.83 - 2vi) / 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?
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.

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 :)

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?

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

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

PhanthomJay
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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?
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!

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.

Finished it. thank you phanthomjay