Solving a Quick Physics Problem: Finding Speed, Deceleration, and Distance

[Rade1990]

Homework Statement

A rock is pushed from rest across a sheet of ice, accelerating at 2.0 m/s2 over a distance of 20 cm. The rock then slides at a constant speed for 15.0 s until it reaches a rough section that causes the rock to stop in 5.5 s. (Assume two significant digits.)

1. What is the speed of the rock when it reaches the rough section?
2. At what rate does it slow down once it reaches the rough section?
3. What is the total distance that the rock slides?
v1=0ms
v2=30m=3000cms
t=15s
d=20cm
a=2.0m.s2

Homework Equations

V2=v1+a (change in t)

The Attempt at a Solution

v2=0+2.0(15)
v2=3000cm =30m

Thats all i understood could some1 show me step by step how to do a b c pleaseee!
thank you!

 

[tiny-tim]

Hi Rade1990!

V2=v1+a (change in t)

But you don't know the change in t … you only know the change in x.

v2=0+2.0(15)

The 15 seconds is when there's no change in v.

 

[Rade1990]

i already did that:S?

 

[kkrizka]

I suggest that you try to break it up into three sections. The accelerating section, the non-accelerating section and finally the rough (de)accelerating section. First step is to find the speed that the rock gains as it is pushed in the first section.

As tiny-tim pointed out, you do not know the time that this happened for so you cannot use the equations listed. I suggest that you write down all the forms of the kinematic equations, what you know and what you are looking for, and based on that find the final speed.

After you find this speed, I'll give you a hint on what to do next.

 

[hamd22]

I did this problem and got the speed for section one to 0.4m/s and i did not know how the find the next speed of section two we only given the time. thks

 

[tiny-tim]

Welcome to PF!

I did this problem and got the speed for section one to 0.4m/s …

Hi hamd22! Welcome to PF!

0.4?

Show us what formula you used.

 

[hamd22]

sorry! i used v2^2=v1^2+2ad then i got the speed to be 0.8m/s for the first section now how do i find the speed for the next section we only have t=15s. thks

 

[tiny-tim]

sorry! i used v2^2=v1^2+2ad then i got the speed to be o.8m/s for the first section now how do i find the speed for the next section we only have t=15s. thks

No … its √(0.8), isn't it?

For the next section, you need an equation with v and a and t.

 

[hamd22]

No … its √(0.8), isn't it?

the √(0.8) is 0.8, so do we use this as initial v and we have time too.Also we need a, is a zero? how do i find them? thks