An object is given an unknown initial velocity up a ramp on which its acceleration is known to have magnitude 13 cm/s^2. It reaches a maximum distance of 81.38 cm up the ramp before rolling back down.
* What is its initial velocity?
* How many seconds after the initial impetus does the object pass a point 6.5 cm up the ramp from its initial position (give all possible answers)?
For this problem we know the following:
a = 13 cm/s^2
We know that | a | = 13 cm/s^2; a can be + 13 cm/s^2 or - 13 cm/s^2.
`ds = 81.38 cm
vf = 0 (when it reaches the top of the ramp it'll stop briefly)
I used the following equation to solve for v0
vf^2 = v0^2 + 2a`ds
Solving for v0, and substituting 0 for vf we get
v0^2 = -2a`ds
v0^2 = -2(13cm/s^2)(81.38cm)
v0^2 = -2115.88
You can't take the square root of a negative--at least if you need a real-number solution.
The signs of the acceleration and the displacement are in fact opposite (one is up the incline, the other down), so -2 a `ds actually turns out to be positive.
The solutions of v0^2 = 2116 cm^2/s^2 are + 45.99 cm/s and - 45.99 cm/s. You use the solution that fits the situation, and discard the one that doesn't.
v0 = -45.99 cm/s
For `dt I used the following:
vf = v0 + a*`dt
`dt = vf-v0/a
`dt = 0 + 45.99 cm/s/13 cm/s^2
`dt = 3.54 s
It took 3.54 seconds for the object to be pushed up the ramp 81.38 cm.
How long does it take it to reach 6.5 cm?
For this, I found the vAve approximately 23 cm/s, then divided the 6.4 cm by the vAve to find the # of seconds it would take to travel that distance.
You can't divide vAve for the entire ramp into 6.4 cm. The average velocity for the whole ramp is not equal to the average velocity for the 6.4 cm.
6.4 cm/23cm/s = .28 seconds
I also could do this another way, divide the 3.54 seconds by the 81.38 cm to find out how long it took to travel 1cm.
3.54/81.38 = .0435 s/cm then multiply this by 6.4 cm = .28 s
This would also assume that vAve for the 6.4 cm is equal to vAve for the entire ramp.
Another way to solve this would also be to divide the 81.38cm by 6.4cm, this would result in 12.7, and then divide the 3.54s by the 12.7, and you also get: .28s
This method implicitly uses the same assumption as the others.
One correct way to find the time to travel 6.5 cm would be to find the velocity after traveling 6.5 cm, starting at 45.99 cm/s, using vf^2 = v0^2 + 2 a `ds and being careful about the signs of a and `ds.
Then you can average v0 and vf for this interval, and use the average velocity to find `dt.
Another way would be to solve the third equation of motion `ds = v0 `dt + .5 a `dt^2 for `dt. This equation is quadratic in `dt and if you're careful it works, but it's easier to use the 4th equation to get vf then use vAve.
You did a good job here, but you did have some errors. Be sure to see my notes and let me know if you have questions.