- #1
- 34
- 1
Homework Statement
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.
a) What is the magnitude of the average velocity of the sled during each of the 2.00-s intervals after passing the 14.4-m point?
b) What is the acceleration of the sled?
c) What is the speed of the sled when it passes the 14.4-m point?
d) How much time did it take to go from the top to the 14.4-m point?
e) How far did the sled go during the first second after passing the 14.4-m point?
Homework Equations
x=x_0+v_0*t+1/2at^2
Other kinematic equations
The Attempt at a Solution
So the first thing I did was drew a picture with labels to get an idea of what the problem looked like to get a better idea of it.
[Part A] asks for the average velocity of each 2 second interval so I found those by taking their distance between each and dividing it by the time, so I got 5.6m/s, 7.2m/s and 8.8m/s.
[Part B] asks for acceleration and because it is constant I can take any two points and find the acceleration that way so I took 7.2m/s and 5.6m/s and divided that by 2s to get an acceleration of .80m/s^2.
[Part C] asks for the velocity at the 14.4m point. This one I'm having issues with. I'm sure it's a small mistake I'm overlooking but plugging it into the standard equation 25.6m=14.4m+v_0*2s+.5*.80m/s^2*2^2 yields 4.8m/s. Putting it into the v_0=v_1-at equation however, using v_1=5.6m/s, a=.80m/s^2 and t=2s gives me 4.0m/s. I'm confused as to why these are not the same. If I can get this issue worked out I'll be able to work the other questions easily.
If I've done anything wrong in A or B please point it out to me, thanks.