Grade 11 Motion Question -- An object is pushed from rest across a sheet of ice

[aatari]

I have solved the question below and was wondering someone can have a look at it and give some feedback, please.

Homework Statement

1. 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.

a. What is the speed of the object when it reaches the rough section?
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

v_f = v_i+a.t
d = vi.t + 1/2 a.t^2
v_f² = v_i² +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.

 

[kuruman]

Hi aatari and welcome to PF.



Parts (a) and (b) are correct. Over the middle interval of 4.0 s the speed is constant, i.e. there is no acceleration. How much distance does the object cover in 4.0 s if its speed is 2.8 m/s?

 

[aatari]

uch d

I see what you mean. So if that's the case, meaning the speed is constant and there is no acceleration, do we then say that the accelration is 0? Because if we do that then we are left with speed and time and if we multiply we get 11.2.

 

[kuruman]

That is correct. You can think of acceleration as the "speed of the speed". An object's position stays the same when its speed is zero. Likewise, an object's velocity stays the same when its acceleration is zero.

 

[aatari]

So for the third part where the object reaches the rough surface and stops after 2.5 seconds. Is it correct then that I have the distance of 3.5 m and if I combine all the distances, I end up with 15.5m.

 

[kuruman]

That would be correct.

 

[Anke]

What did you do to get 2.8 m/s?

 

[DaveC426913]

What did you do to get 2.8 m/s?

Yeah.

It's not easy to get v, given s instead of t. I'm too lazy to do the savtu substitutions, but pretty sure it can be done.

[EDIT] Ah!

v=√(u²+2as)

https://www.calculatorsoup.com/calculators/physics/velocity-calculator-vuas.php(The final v = 2.828427... looks familiar! It's simply 2√2 - probably not a coincidence.)

 

[berkeman]

What did you do to get 2.8 m/s?

I subtracted 2014.2 from the year 2017 to get that answer...