course Phy 121
I don't know how to do this problem really. I think I pulled quite a bit of hair out in the course of working it. Have we done any problems like this one? I only knew one way to find the initial velocity when I was only given `dt, `ds, and a. Could you please explain to me how I was supposed to find other possible initial velocities? Thank you!
An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude 10 cm/s^2. .123 seconds later it passes a point 6.6 cm up the ramp from its initial position. ?What are its possible initial velocities, and what is a possible scenario for each?
Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.
What is the maximum distance the object travels up the ramp?
a = 10cm/s^2, `dt = 0.123s, and `ds = 6.6cm
The only information we can find with these variables is change in velocity and average velocity.
vAve = `ds/`dt = 6.6cm/0.123s = 53.659 cm/s
`dv = a`dt = (10cm/s^2)(0.123s) = 1.23 cm/s
I had wondered if vo could equal zero but after double checking with the information above, I see that the initial velocity could not be zero. If vo = 0, vf = `dv = 1.23cm/s. Then the change in velocity would be 0.615 That would be the average velocity, but either way it certainly won't be so
. We know this is not true. So basically, the change in velocity and average velocity are dead ends.
The only equation we can use for this problem and find any useful information is `ds = (vo)(`dt) + 1/2(a)(`dt)^2
So we rearrange it to solve for vo:
vo = [`ds - 1/2(a)(`dt)^2 ]/`dt
vo = [6.6cm - 1/2(10cm/s^2)(.123s)^2]/0.123s
vo = 53.044 cm/s
Good. Note that you could have found this from your previous work. vAve = 53.659 cm/s and `dv = 1.23cm/s. Since vAve is halfway between v0 and vf, half the `dv occurs between v0 and vAve. Half the `dv is .615 cm/s; subtracting this from vAve we get v0 = 53.659 cm/s - .615 cm/s = 53.044 cm/s.
Either way you are doing very well to find v0 on this challenging problem.
There is no way to find the distance traveled on the ramp if we do not know the time interval or some other combination of variables. Since 53.044 cm/s is the only initial velocity we can find, we assume that the distance the object travels is the distance that corresponds to this velocity, acceleration, and time interval. Therefore, the maximum distance traveled is 6.6 centimeters.
To find the maximum distance you can use v0 and a, along with the knowledge that at maximum distance we have vf = 0. This gives you three variables and you can solve the problem.
Now note one more detail. The ball is accelerating down the ramp and moving up. The magnitude of the acceleration is 10 cm/s^2. If you take up as positive, then a must be negative. You would then get v0 = 54.274 cm/s, I believe, and the absolutely correct answers would follow from this.