- #1
aatari
- 73
- 3
I have solved the question below and was wondering someone can have a look at it and give some feedback, please.
b. At what rate does the object slow down once it reaches the rough section?
c. What total distance does the object slide throughout its entire trip?
vf = vi+a.t
d = vi.t + 1/2 a.t^2
vf2 = vi2 +2.a.d[/B]
[/B]
For part a using the given accelration and distance I found the velocity to be 2.8 m/s
However, for part b and c I am confused and although I solved to get the answer but I feel my answers are not correct.
b. So when the object reaches the rough section its velocity is still 2.8 m/s, and it stops after 2.5 s So using the first equation above I got the accelration to be -1.12 m/s2 by dividing 2.8 with 2.5s.
c. Fort this last question I divided the above scenario into three parts, first part the distance is 0.8m (its given). For the second part only time is give, which is 4.0 s but I know the velocity from question a, so using the time and velocity I figured out the acceleration which was 0.71 and then using all this information I found the distance of the second part to be 16.96 m, which seems high to me. Finally, to find the distance for the last part, only time is given, 2.5s. But from question b I know that the accelration is -1.12. So then I found the velocity with this information and finally using the third equation above I found the distance for this part to be 3.5m.
Adding all three parts I got the total distance of 21.26m.
Homework Statement
- An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80.0 cm. The object then slides with a constant speed for 4.0 s until it reaches a rough section which causes it to stop in 2.5 s.
b. At what rate does the object slow down once it reaches the rough section?
c. What total distance does the object slide throughout its entire trip?
Homework Equations
vf = vi+a.t
d = vi.t + 1/2 a.t^2
vf2 = vi2 +2.a.d[/B]
The Attempt at a Solution
[/B]
For part a using the given accelration and distance I found the velocity to be 2.8 m/s
However, for part b and c I am confused and although I solved to get the answer but I feel my answers are not correct.
b. So when the object reaches the rough section its velocity is still 2.8 m/s, and it stops after 2.5 s So using the first equation above I got the accelration to be -1.12 m/s2 by dividing 2.8 with 2.5s.
c. Fort this last question I divided the above scenario into three parts, first part the distance is 0.8m (its given). For the second part only time is give, which is 4.0 s but I know the velocity from question a, so using the time and velocity I figured out the acceleration which was 0.71 and then using all this information I found the distance of the second part to be 16.96 m, which seems high to me. Finally, to find the distance for the last part, only time is given, 2.5s. But from question b I know that the accelration is -1.12. So then I found the velocity with this information and finally using the third equation above I found the distance for this part to be 3.5m.
Adding all three parts I got the total distance of 21.26m.