How to find the difference between two station?

[louis676]

Homework Statement

A train left the station by accelerating at 5.3 m/s ² for 5 seconds. It then continued at constant speed for 64 seconds. As it approached the next station, it slowed to a stop at 4.2 m/s ².

What is the distance between the 2 stations?

Homework Equations

Δx = V0Δt + .5a(Δt)2
I assume is this forumula.

The Attempt at a Solution

I don't know how to approach this question

 

[berkeman]

Homework Statement

A train left the station by accelerating at 5.3 m/s ² for 5 seconds. It then continued at constant speed for 64 seconds. As it approached the next station, it slowed to a stop at 4.2 m/s ².

What is the distance between the 2 stations?

Homework Equations

Δx = V0Δt + .5a(Δt)2
I assume is this forumula.

The Attempt at a Solution

I don't know how to approach this question

Yes, that is the correct equation.

Now write the 3 versions of that equations for the 3 parts of the trip, and add up the Δx values that you get for each of the 3 parts of the trip.

 

[louis676]

For the three version you mentioned, are the following the right approaches?

If this is the right way, how do I find V_o
Δx = V0Δt + .5a(Δt)2
Δx =?
V₀ =
Δt = 5 s
A = 5.3 m/s ²

Same thing, if this is the right approach, how do I find V_o
Δx = V₀Δt + .5a(Δt)²
Δx =?
V₀ =?
Δt = 64 s
A = 5.3 m/s ²

The information provided for this one is too vague, not certain did I plug in the variable correctly.
Δx = V₀Δt + .5a(Δt)²
Δx =?
V₀ =?
Δt =
A = 4.2 m/s ²

 

[azizlwl]

Another approach i usually do is to draw a velocity-time the graph.
At t=0, v=0
(a).Then for 5 seconds, the velocity increases from 0 to V₂
The gradient at between t=0 to t=5sec is 5.3. dv/dt=5.3
(b).For another 64 s level off at V₂.
(c). Next the graph has negative gradient of 4.2 till v=0.

Remember area under the graph(velocity x time) is the displacement.

 

[louis676]

I'm not really good with graph lol..
I know using graph is easier to find answer, but my teacher insist to use equation.

 

[azizlwl]

You have a correct equation for constant acceleration.
But you do not have the values for the variables.
From the graph you can deduce the values of those variable.
Maybe you try to find out how the equation is derived from.

 

[berkeman]

For the three version you mentioned, are the following the right approaches?

If this is the right way, how do I find V_o
Δx = V0Δt + .5a(Δt)2
Δx =?
V₀ =
Δt = 5 s
A = 5.3 m/s ²

Same thing, if this is the right approach, how do I find V_o
Δx = V₀Δt + .5a(Δt)²
Δx =?
V₀ =?
Δt = 64 s
A = 5.3 m/s ²

The information provided for this one is too vague, not certain did I plug in the variable correctly.
Δx = V₀Δt + .5a(Δt)²
Δx =?
V₀ =?
Δt =
A = 4.2 m/s ²

For the first part, the train starts sitting in the station. What is Vo there?

At the end of the first part, the train has some velocity. What is it?

What is the velocity at the start of the 2nd part? Does the velocity change during the 2nd part?

And keep going on the rest...

 

[louis676]

For the first part, the train starts sitting in the station. What is Vo there?

At the end of the first part, the train has some velocity. What is it?

What is the velocity at the start of the 2nd part? Does the velocity change during the 2nd part?

And keep going on the rest...

Is this correct?

1s part
Δx = V0Δt + .5a (Δt) ²
Δx =?
V0 = 0 m/s
Δt = 5 s
A = 5.3 m/s ²

Δx = 0 m/s (5s) + .5 (-9.8m/s ²) (5s) ²
Δx = 122.5


2nd part
Δx = V0Δt + .5a (Δt) ²
Δx =122.5
V0 =?
Δt = 64 s
A = 5.3 m/s ²

Δx = V0Δt + .5a (Δt) ²
122.5 m= VO (64s) + .5 (-9.8 m/s ²) (64s) ²
122.5 m= VO (64s) – 4.9 m/s ² (64s) ²
122.5 m= VO (64s) – 20070.4 m
20192.9 m= VO 64s * Divide 64 both sides
315.5 m/s = Vo


3rd Part
Δx = V0Δt + .5a (Δt) ²
Δx =?
V0 =315.5 m/s
Δt =?
A = 4.2 m/s ²

Δx = 315.5 m/s (Δt) + .5(-4.2 m/s ²) (Δt) ²
Δx = 315.5 m/s (Δt) - 2.1 m/s ² (Δt) ²
I assume we need to use quadratic equation.
Ax ² = -2.1
Bx = 315.5
Cx = 1

 

[berkeman]

The first part is correct, but also use the velocity equation to tell you what the velocity is at the end of the first part. You need that as the Vo for the second part...

 

[louis676]

For the 1st part, I'm not really sure is it correct.
The acceleration I plug in is -9.8 m/s ^ 2, but shouldn't it be -5.3m/s ^ 2?

Also for the 1st part, the answer I got is negative. But is correct to have a negative height?

 

[berkeman]

For the 1st part, I'm not really sure is it correct.
The acceleration I plug in is -9.8 m/s ^ 2, but shouldn't it be -5.3m/s ^ 2?

Also for the 1st part, the answer I got is negative. But is correct to have a negative height?

No, -9.8m/s^2 is for vertical motion questions where gravity is affecting the motion. This is a train moving on a horizontal track.

You correctly listed this in the first part: A = 5.3 m/s 2
I didn't notice that you plugged in a different acceleration into the equation for the first part. Please fix that, and find the distance for the first part. Also write the velocity equation for the first part, and find the final velocity of the train for the first part. Then proceed...

 

[louis676]

No, -9.8m/s^2 is for vertical motion questions where gravity is affecting the motion. This is a train moving on a horizontal track.

You correctly listed this in the first part: A = 5.3 m/s 2
I didn't notice that you plugged in a different acceleration into the equation for the first part. Please fix that, and find the distance for the first part. Also write the velocity equation for the first part, and find the final velocity of the train for the first part. Then proceed...

To find distance:
Δx = V0Δt + .5a (Δt) 2
Δx =?
V0 = 0 m/s
Δt = 5 s
A = 5.3 m/s 2
Δx = 0 m/s (5s) + .5 (-5.3 m/s 2) (5s) 2
Δx = 0 m + 2.65m/s 2 (25s 2)
Δx = 0 m + 2.65m/s 2 (25s 2)
Δx = 0 m + 66.25 m
Δx = 66.25 m

I use this method to find Vo, however, Vo comes out to be 0.
Δx = V0Δt + .5a (Δt) 2
66.25 = Vo (5s) + 2.65 m/s ^2 (25s) ^2
66.25 = Vo (5s) + 66.25 m

Which method should I use to find Vo?

 

[berkeman]

This is the equation for distance in terms of initial velocity and acceleration (actually there is usually an initial position term too, but that is zero when the train starts from the station):

Δx = V0Δt + .5a (Δt) 2

There is a simiar equation that relates the velocity v(t) to the initial velocity and the acceleration... What is that equation? What does it give you for an answer for the final velocity in the first part?

 

[louis676]

I believe the equation to find Initial velocity is,
Vo = V - AT
But I forgot what is V.
From what I remember, I think V = Vf - Vo isn't it? So if I don't know neither Vo and Vf, how can I calculate V?

And the equation for final velocity is (Vf)^2 = (Vo)^2 + 2aΔx.

 

[berkeman]

I believe the equation to find Initial velocity is,
Vo = V - AT
But I forgot what is V.
From what I remember, I think V = Vf - Vo isn't it? So if I don't know neither Vo and Vf, how can I calculate V?

And the equation for final velocity is (Vf)^2 = (Vo)^2 + 2aΔx.

The velocity equation is equation [1] in the section at this page about Uniform Acceleration (about 1/3 of the way down the page):

http://en.wikipedia.org/wiki/Equations_of_motion

.

 

[louis676]

Alright Thanks.

 

[louis676]

I know why I need to find Initial velocity, but why do I need to find final velocity?

 

[berkeman]

I know why I need to find Initial velocity, but why do I need to find final velocity?

The final velocity for segment 1 is the initial velocity for segment 2, and so on...

Makes sense?

 

[louis676]

Yes I understand..
Thank you!

 

[Steely Dan]

Please keep your acceleration signs consistent. If the speed is increasing the acceleration is positive, if it is constant then the acceleration is zero, and if the speed is decreasing then the acceleration is negative.