A train travelling between stations

[Final_HB]

Hey. This is a little tricky

 

The question goes:

A train travels from rest at station A to rest again at station B in a time of 4 minutes and a distance of 9km. The greatest speed attained is 60m/s, and both the acceleration and decceleration are uniform.

 

Find how far the train goes at full speed?

 

I know what I need to do to get the answer, but I dont have enough info for it.

 

-Find the time at a constant speed (T_c)

-Find distance using the formula: S=UT_c

 

To find T_c, I need either:

-Distance travelled during deceleration/acceleration.

-Time to accelerate from rest to 60m/s.

-Time to decelerate from 60m/s to rest.

 

Get any of those four, some maths wizardry and im golden

 

But i cant do anything with the info given.

Am I missing a beat?

[studiot]

You have enough information to solve this.

 

I suggest you draw a velocity-time graph. The distance is the area under the graph.

 

What can you tell me about the acceleration periods and deceleration periods?

[Final_HB]

I assumed I did Yup... drawn. Wouldnt get me an A in art, but its here.

The graph divides up into three shapes, Two triangles, and a retangle.

 

The retangle is the area we are looking for, and the triangles are the acceleration/ deceleration parts.

 

EDIT: CallingT_a the time accelerating and T_d the time decelerating, we can equate the distance to the following:

60T_a/2 + 60T_d/2 + 60t_c =9000

which reduces to,

T_a +T_d +2T_c =300

Edited August 14, 2013 by Final_HB

[studiot]

 

The graph divides up into three shapes, Two triangles, and a retangle.

 

 

Since you didn't answer my question about the acceleration and deceleration before I will ask it again (hint hint)

 

What can you say about the two triangles?

 

Now can you write (and post here)

 

1) an equation connecting the information given (distance) and your graph?

 

2) an equation connecting the times on your graph and the information given?

 

EDIT: CallingT_a the time accelerating and T_d the time decelerating, we can equate the distance to the following:

60T_a/2 + 60T_d/2 + 60t_c =9000

which reduces to,

T_a +T_d +2T_c =300

 

 

So what can you say about T_a and T_d ?

Edited August 14, 2013 by studiot

[Final_HB]

Sorry. The area of the triangles is the distance travelled while accelerating/decelerating.

Uniform...

Does that mean they are equal, or constant?

If they are equal, the distance travelled must be the same... so T_a = T_d

[studiot]

I haven't seen the original question, but the numbers work out nicely if they are equal, and you do not have enough information if they are not.

 

Can you also post my second equation?

[Final_HB]

The original post has the question word for word

T_{a +}T_d +T_c =240

 

From what I said in Post #5,

2T_a + T_c =240

 

So... simultaneous equations and solve for T_c ?

[studiot]

 

So... simultaneous equations and solve for T_c ?

 

Yes.

 

You will have to take up the poor wording of the original with your tutors.

[Final_HB]

Thank you!!

 

Agreed :/

 

thanks all the same

[imatfaal]

Dunno if you are still around Final_HB

 

the question is fine. Your velocity time graph gives you a trapezoid. The Area of a trapezoid is

 

Area =Height * (long parallel side + short parallel side)/2

 

Area = total distance

Height = max speed

long parallel speed = total time

short parallel speed = time at top speed -> unknown

 

only one unknown - ie can be solved without making assumptions about acceleration. you now know time at full speed and as speed x time equals distance you know that as well.

[studiot]

Good thinking, I missed that, imatfaal.

 

[imatfaal]

Geometrically it is obvious - but in a physics scenario it is far from obvious. To be honest I am still slightly struggling to see intuitively why it must be so

[studiot]

I had a ngging feeling that there was something about the area, but was too lazy to work it out.

 

Needless to say I have since worked it out properly and you are quite correct.

[Final_HB]

Hey just saw the post.

Wow, okay that works to Im just used to splitting the graphs into easy geometric shapes... squares, retangles, triangles ya know... Because thats how we were shown

 

In all honesty, I didnt realize that there an area formula for traps

 

Thank you, imatfaal and studiot

Edited August 15, 2013 by Final_HB