HI, I have a question where a roller-coaster does 55m in 3s and we are asked to calculate its acceleration assuming it's vertical and constant. Other than that, there is no other information given, no initial velocity, no final velocity. So can you find the acceleration of an object with only distance and time given?

52 minutes ago, DODOma said:

HI, I have a question where a roller-coaster does 55m in 3s and we are asked to calculate its acceleration assuming it's vertical and constant. Other than that, there is no other information given, no initial velocity, no final velocity. So can you find the acceleration of an object with only distance and time given?

Since this is homework what have you done towards It ?

Have you read the question properly because as you have stated it a solution is not possible.

You say neither initial nor final velocities are supplied  so what equation were you thinking of using ?

However I suspect there is actually more information supplied indirectly.

Edited September 17, 20232 yr by studiot

Since roller coasters are not powered and the question says the coaster is falling vertically the acceleration should be easy to determine, think about it.

18 minutes ago, Bufofrog said:

Since roller coasters are not powered and the question says the coaster is falling vertically the acceleration should be easy to determine, think about it.

Must be awfully powerful gravity at this theme park.   110/9  = 12.22

It says It's going up vertically.

19 minutes ago, Bufofrog said:

Since roller coasters are not powered and the question says the coaster is falling vertically the acceleration should be easy to determine, think about it.

It says it's going up not coming down. In that case it would be easier a=g.

48 minutes ago, studiot said:

Since this is homework what have you done towards It ?

Have you read the question properly because as you have stated it a solution is not possible.

You say neither initial nor final velocities are supplied  so what equation were you thinking of using ?

However I suspect there is actually more information supplied indirectly.

I used this equation but there are 2 unknown values couldn't proceed, x= at^2/2 + vt + x

simplifies to a=(110 - 6v)/ 9. v is initial velocity. Because initial velocity is not given anywhere in the question, i got stuck here. So i thought i should ask a question on this forum because i never knew you could find acceleration with just distance and time given. 😅

43 minutes ago, DODOma said:

I used this equation but there are 2 unknown values couldn't proceed, x= at^2/2 + vt + x

simplifies to a=(110 - 6v)/ 9. v is initial velocity. Because initial velocity is not given anywhere in the question, i got stuck here. So i thought i should ask a question on this forum because i never knew you could find acceleration with just distance and time given. 

Well you haven't post the original question verbatim.

 

 

1 hour ago, DODOma said:

It says it's going up not coming down. In that case it would be easier a=g.

Am I missing something? Going up or down, the acceleration would still be +g in the downward direction. 

Or minus g in the upwards direction. Unless you factor in drag and friction.

43 minutes ago, studiot said:

Well you haven't post the original question verbatim.

 

 

It says the rollercoaster climbs 55 meters in 3 seconds. The question is: calculate the acceleration assuming it's constant and vertical. What would happen to the acceleration if the if the time taken to climb was 4 seconds. That's the 1st question.

Just now, mistermack said:

Am I missing something? Going up or down, the acceleration would still be +g in the downward direction. 

Or minus g in the upwards direction. Unless you factor in drag and friction.

It's going up.

2 minutes ago, DODOma said:

It's going up.

Why does that make a difference?

6 minutes ago, DODOma said:

It says the rollercoaster climbs 55 meters in 3 seconds. The question is: calculate the acceleration assuming it's constant and vertical. What would happen to the acceleration if the if the time taken to climb was 4 seconds. That's the 1st question.

Can you post the entire question, verbatim, please?

33 minutes ago, swansont said:

Can you post the entire question, verbatim, please?

The question is not in English so i used google translate to translate the question

28 minutes ago, DODOma said:

The question is not in English so i used google translate to translate the question

OK, it’s not identified as a roller coaster, and the implication is that you are starting from rest.

You can use s = 1/2 at^2

3 minutes ago, swansont said:

OK, it’s not identified as a roller coaster

That is correct.

(it is a vertical drop*, the information in the question is enough to find images and videos of the ride on the net @DODOma )

*  "drop tower" or "big drop" https://en.wikipedia.org/wiki/Drop_tower

We have both French and Dutch Speakers here so please post the original question entirely.

23 hours ago, studiot said:

Must be awfully powerful gravity at this theme park.   110/9  = 12.22

There was no initial velocity specified and I incorrectly assumed (since it was not stated) that the vertical travel was a drop, but if my assumption had been correct an initial velocity of 3.63 m/s would yield a travel distance of 55m in 3 seconds with the acceleration of 9.8 m/s^2.