course Phy 231
***A ball starting from rest rolls 11 cm down an incline on which its acceleration is 27 cm/s2, then onto a second incline 44 cm long on which its acceleration is 9 cm/s2. How much time does it spend on each incline?***We can use one of the equations of motion with uniform acceleration to solve the time for the first incline.
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.
`ds = Vo`dt + 0.5a`dt^2
11cm = 0`dt + 0.5(27cm/s^2)(`dt^2)
11cm = (13.5cm/s^2)(`dt^2)
`dt^2 = 0.815s^2
`dt = 0.903s
The ball was on the first incline for 0.903 seconds.
The second incline is done the same way but with a little more work. It's initial velocity is no longer zero so we have to find the final velocity at the end of the first incline. This will be the initial velocity for the second incline.
To find the final velocity for the first incline (initial velocity for the 2nd incline):
Vf^2 = Vo^2 + 2a`ds
Vf^2 = 0 + 2(27cm/s^2)(11cm)
Vf^2 = 594cm^2/s^2
Vf = 24.4cm/s
Therefore 24.4cm/s is the final velocity for the first interval and the initial velocity for the second interval. Now we use the equation for motion with uniform acceleration again:
`ds = Vo`dt + 0.5a`dt^2
44cm = 24.4cm/s*`dt + 0.5(9cm/s^2)`dt^2
44cm = 24.4cm/s*`dt + 4.5cm/s^2`dt^2
44cm = `dt(4.5cm/s*`dt + 24.4cm/s)
4.5`dt + 24.4 = 44
4.5`dt = 19.6
`dt = 4.36s"
Good work, with only one exception.
You did not solve the last equation correctly for `dt. That equation is quadratic in `dt and would be solved using the quadratic formula.
You are generally better off avoiding the issue of the quadratic. Solve the fourth equation for vf, then use that with your v0 to find vAve. Divide `ds by vAve.