Solving the Physics of an Object's Slide

[barthayn]

Homework Statement

An object is pushed along a rough horizontal surface and released. it slides for 10.0s before coming to rest and travels a distance of 0.2m during the last 1.0s of its slide. Assuming the acceleration to be uniform throughout
a) How fast was the object traveling upon release?
b) How fast was the object traveling when it reached the half way position in its slide?

v_f = 0m/s
v_i = ?
d₁ = ?
d₂ = 0.2m
a = ?
t = 10s

Homework Equations

a = Δv/Δt
Δd = vΔt = ((V_i+V_f)/2)Δt
v_f = Δt-1/2a(Δt)²
Δd = v_fΔt-1/2a(Δt)²

The Attempt at a Solution

a = -0.4/1
a = -0.4m/s²

Δd = (0.4(100))/2
Δd = 40/2
Δd = 20 m

v_i² = 0-(-0.2)(20.2)
v_i² = 4.04m²/s²
v_i = 2.009m/s

Therefore the object was traveling at 2.009 m/s when first released.

For part b, I do not know where to begin. The acceleration is -0.4m/s², as founded in the first part. But I do not know how to apply the information because I get a negative answer for the final velocity for the half way point.

 

[Redbelly98]

Homework Statement

An object is pushed along a rough horizontal surface and released. it slides for 10.0s before coming to rest and travels a distance of 0.2m during the last 1.0s of its slide. Assuming the acceleration to be uniform throughout
a) How fast was the object traveling upon release?
b) How fast was the object traveling when it reached the half way position in its slide?

v_f = 0m/s
v_i = ?
d₁ = ?
d₂ = 0.2m
a = ?
t = 10s

Homework Equations

a = Δv/Δt
Δd = vΔt = ((V_i+V_f)/2)Δt
v_f = Δt-1/2a(Δt)²
Δd = v_fΔt-1/2a(Δt)²

The Attempt at a Solution

a = -0.4/1
a = -0.4m/s²

Δd = (0.4(100))/2
Δd = 40/2
Δd = 20 m

Your notation is a little on the sloppy side, but you are basically okay up to here.

v_i² = 0-(-0.2)(20.2)
v_i² = 4.04m²/s²
v_i = 2.009m/s

Where does this come from? It seems wrong to me.

Therefore the object was traveling at 2.009 m/s when first released.

I got a different answer. Can you explain the stuff beginning with "v_i²=0-(-0.2)(20.2)" please?

For part b, I do not know where to begin. The acceleration is -0.4m/s², as founded in the first part. But I do not know how to apply the information because I get a negative answer for the final velocity for the half way point.

A negative answer is a problem. Can you show your work, then I can help figure out what is wrong.

 

[barthayn]

Your notation is a little on the sloppy side, but you are basically okay up to here.


Where does this come from? It seems wrong to me.


I got a different answer. Can you explain the stuff beginning with "v_i²=0-(-0.2)(20.2)" please?


A negative answer is a problem. Can you show your work, then I can help figure out what is wrong.

v_i²=0-(-0.2)(20.2) is from the acceleration found, and the total distance traveled.

The final answer I got 2.009m/s. For b) I got 2.04m/s. To get that answer I did this:

Divided the total distance and time by two.

v_f = (2d+a(t)²)/t
v_f = (20.2-10)/5
v_f = 10.2=5
v_f = 2.04 m/s

 

[Redbelly98]

v_i²=0-(-0.2)(20.2) is from the acceleration found, and the total distance traveled.

The acceleration is -0.4 m/s², but that value does not appear anywhere in your equation.
The distance traveled is 20.0 m, but you have 20.2 in your equation.
If you're using the equation I think you're using, check whether it has a factor of 1/2 or 2. OR you could use the simpler a=Δv/Δt that you listed as one of the Relevant Equations.

The final answer I got 2.009m/s. For b) I got 2.04m/s. To get that answer I did this:

Divided the total distance and time by two.

Actually, it is just the distance that is divided by two, not the time:

b) How fast was the object traveling when it reached the half way position in its slide?

"half way position" refers to the distance.

 

[barthayn]

The acceleration is -0.4 m/s², but that value does not appear anywhere in your equation.
The distance traveled is 20.0 m, but you have 20.2 in your equation.
If you're using the equation I think you're using, check whether it has a factor of 1/2 or 2. OR you could use the simpler a=Δv/Δt that you listed as one of the Relevant Equations.


Actually, it is just the distance that is divided by two, not the time:

"half way position" refers to the distance.

You get the 0.2m plus from the question given. Also, in my notes, I got -0.4 m/s². So I mistyped the number. The rest is correct.

How would I get the time to get it to reach the half way point. It states that the acceleration is uniform.

 

[Redbelly98]

You get the 0.2m plus from the question given.

According to the problem statement, the object comes to rest in 10.0 s.
You correctly calculated that it moves 20 m in those 10 seconds. At that point it is at rest, so it cannot travel an additional 0.2 m. Instead, those 0.2 m comprise the final part of the 20 m total distance (and the final 1.0 s of the 10.0 s total).

Also, in my notes, I got -0.4 m/s². So I mistyped the number. The rest is correct.

It is not correct, since you (1) used the mistyped -0.2 acceleration value to do the calculation, and (2) used a formula that I think is incorrect.

What is the general formula you used, before you substituted values into it? I mean the formula that has vi²=______? My further assistance is contingent upon your providing this information.

How would I get the time to get it to reach the half way point. It states that the acceleration is uniform.

You have acceleration and distance. Once part (a) is done correctly, you'll have initial velocity. Three quantities are sufficient to solve any 1-d uniform acceleration problem, you just need to find and use an equation that has a, Δd, and v_i.

 

[barthayn]

According to the problem statement, the object comes to rest in 10.0 s.
You correctly calculated that it moves 20 m in those 10 seconds. At that point it is at rest, so it cannot travel an additional 0.2 m. Instead, those 0.2 m comprise the final part of the 20 m total distance (and the final 1.0 s of the 10.0 s total).It is not correct, since you (1) used the mistyped -0.2 acceleration value to do the calculation, and (2) used a formula that I think is incorrect.

What is the general formula you used, before you substituted values into it? I mean the formula that has vi²=______? My further assistance is contingent upon your providing this information.You have acceleration and distance. Once part (a) is done correctly, you'll have initial velocity. Three quantities are sufficient to solve any 1-d uniform acceleration problem, you just need to find and use an equation that has a, Δd, and v_i.

The new first velocity I got was 4 m/s. Is this correct? For part b I got 2 m/s as well

 

[Redbelly98]

I agree with the 4 m/s. But it would be 2 m/s at half the total time, not half of the distance.

p.s. logging off for the night, will be back tomorrow.

 

[barthayn]

I agree with the 4 m/s. But it would be 2 m/s at half the total time, not half of the distance.

p.s. logging off for the night, will be back tomorrow.

How would I find out the time from this then? I am completely confused because the negative acceleration is uniform, meaning it is constant?

 

[Redbelly98]

Yes, the acceleration is constant.

You have Δd, a, and v_i. Look through your list of kinematic equations for an equation that has those three terms in it.

 

[barthayn]

Yes, the acceleration is constant.

You have Δd, a, and v_i. Look through your list of kinematic equations for an equation that has those three terms in it.

How would I find the time it took to reach the half way distance then? I do not understand how you can get it mathematically. Wouldn't it just be total time divided by two because it is consent negative acceleration.

 

[Redbelly98]

Not, it would have to be constant velocity for it to go half the total distance in half the total time. Hopefully that is common sense. This is not a constant velocity situation, since acceleration is not zero.

But -- guess what? -- you do not even need to know what that time is anyway. Use the kinematic equation that has Δd, a, v_i (all of which you know) and v_f (which you are trying to find).

Do you, or do you not, have the kinematic equations in your textbook or class notes?

 

[barthayn]

The final answer I got was 2.8 m/s. Is this correct?

 

[Redbelly98]

Yes, looks good to me