How to Determine Acceleration from Distance in Constant Acceleration Problems

[kylenic1997]

Homework Statement

A boulder starts from rest and travels 2.70 m in the first second (from t = 0 s to t = 1 s). How far will it travel in the second second (from t = 1 s to t = 2 s)? Assume the acceleration of the boulder is constant.

Homework Equations

a=(V-Vo)/t
X=Xo+at
v^2-Vo^2=2a(X-Xo)

The Attempt at a Solution

My problem is figuring which equation to use because I keep getting different accelerations depending on which one I use. The first equation gives me a=(2.70m/s)/1s =2.70m/s^2. Second equation gives me a= 2.70m/s, looking back now I guess this answer cannot be it because it is not in m/s^2. Last equation gives me (2.70m/s)^2 /(2.70m *2)= 1.35m/s^2. How can I determine which equation should be used for this problem and for future problems. I have noticed that I can normally determine the other variables well, but I keep getting the acceleration wrong. Thank you!

 

[Simon Bridge]

My problem is figuring which equation to use...

... your problem is that you are trying to remember the equation instead of understanding the physics.

...I keep getting different accelerations depending on which one I use. The first equation gives me a=(2.70m/s)/1s =2.70m/s^2.

How did you get 2.7m/s for this equation?

Second equation gives me a= 2.70m/s, looking back now I guess this answer cannot be it because it is not in m/s^2.

Well done.

Notice that you are not told the speed after 1s, just the distance travelled.
Remember to list what you know, write down the unknown as a ?, then pick the equation that has everything you know and the unknown only.
Note: there are 5 suvaT equations

Better: get used to sketching velocity time graphs.
Example - object accelerates from rest, travels 5m in the first second ... what is the final speed?
The v-t graph is a triangle, the apex is unknown at v, base is 1s long, the area is 5m = v/2 (half base times height) so v=10m/s is the final speed.

 

[kylenic1997]

I thought that velocity= distance traveled/ time traveled. Since it traveled 2.70m/1s I figured it would be 2.70 m/s. From the example you gave me it seems like the last equation of my original post should be correct. For some reason it seems to make more sense that the first equation should be correct, because I keep thinking that the velocity should be 2.70 m/s after the first second.

 

[haruspex]

he first equation gives me a=(2.70m/s)/1s

Where are you getting 2.70m/s from? I don't see that in the provided data,

X=Xo+at

Where x is a distance? 'at' is acceleration times time. That does not give a distance.

 

[kylenic1997]

I got 2.70m/s from (X-Xo)/t. With the info given I got (2.70m-0m)/1s. That gave me 2.70m/s. If that's not how I get the velocity, then I don't know how to. Those are the only formulas my professor has us use.

 

[jbriggs444]

I got 2.70m/s from (X-Xo)/t. With the info given I got (2.70m-0m)/1s. That gave me 2.70m/s. If that's not how I get the velocity, then I don't know how to. Those are the only formulas my professor has us use.

That equation gives you the average velocity. If the boulder is accelerating, its starting velocity will be lower than that and its ending velocity will be higher than that. Knowing an average velocity will not help you calculate acceleration.

So... re-read post #2. What other equations might you consider using?

 

[Simon Bridge]

Hmmm... the second equation in the list post #1 is incorrect anyway.
Best not to rely on memorizing equations.

If you started at velocity v and stayed at that velocity for time t, how far did you travel (whats the equation)?
If you started at 0, and ended at v by constant accelerarion, would you have traveled farther or less far than if you traveled at a constant speed? Can you work out what the equation has to be?

Hint: sketch the v-t graph... displacement is the area under the graph.