A rolling ball, find distance traveled

[rebbe90]

Homework Statement

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 200m during the second 5s of it's motion. How far did it roll during it's first 5s of motion.

Relevant Equations

V = s/t

I figured the best way to do this is to focus on the second half of time. We can use the information there to find acceleration and that should make it fairly simple to find distance traveled in the first 5 seconds. Average speed in 5s-10s I found to be 40m/s.

My problem is that to find the acceleration I think I have to find the position as a function of time and derive it. Stuck on how to make the function.

 

[ergospherical]

The standard expressions of uniform acceleration tell you that the position varies quadratically with time, $x = \frac{1}{2} at^2 \propto t^2$. If you double the time, then by what factor does the position increase?

 

[PeroK]

My problem is that to find the acceleration I think I have to find the position as a function of time and derive it. Stuck on how to make the function.

What about using a spreadsheet and setting a variable equal to a value of acceleration. Then compute the distance in the first $5s$ and in the second $5s$. Tune the variable until the distance in the second $5s$ is $200m$. That would give you the answer for the first $5s$.

Then, try to figure out algebraically what the tuning process is doing.

 

[bob012345]

The standard expressions of uniform acceleration tell you that the position varies quadratically with time, $x = \frac{1}{2} at^2 \propto t^2$. If you double the time, then by what factor does the position increase?

This hint is a great example of how understanding the principles involved will quickly cut through a mass of formulas and get to the right answer fast, especially useful on a test.

 

[kuruman]

I add a geometric suggestion to the others. The distance covered over a time interval is the area under the velocity vs. time curve. Look at the v vs. t graph below. The distance covered in the first 5 s is represented by the area in red. The distance covered in the next 5 s is represented by the area in green. If the green area is 200 m, how big is the red area?