Max Subway Speed & Time b/w Stations, Avg Max Speed w/Stops

[mossfan563]

Homework Statement

(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.54 m/s2 and subway stations are located 816 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?

Homework Equations

velocity & time:
v = v_0 + a t

displacement & time:
x = x_0 + v_0 t + (1/2) a t^2

velocity & displacement:
v^2 = v_0^2 + 2 a \\Delta x

The Attempt at a Solution

I tried the displacement and time equation for part A and it didn't work. I had no idea how to do the other two parts.
How do you start this problem?

 

[tiny-tim]

I tried the displacement and time equation for part A and it didn't work.

Hi mossfan563!

It's difficult to say why it didn't work if you don't show us what you did!

I'll guess for now … did you remember to use only half the distance?

 

[mossfan563]

Hi mossfan563!

It's difficult to say why it didn't work if you don't show us what you did!

I'll guess for now … did you remember to use only half the distance?

Well I simply plugged in the distance between stations as the equation's displacement and the acceleration is given. I have to use half the distance?

 

[tiny-tim]

Well I simply plugged in the distance between stations as the equation's displacement and the acceleration is given. I have to use half the distance?

erm … depends on whether you want people to be able to get on and off without nets!

 

[mossfan563]

Seriously, how do i start this problem?

 

[tiny-tim]

Seriously, how do i start this problem?

You need to accelerate it for 408m at 1.54 m/s², and then decelerate it for the other 408m.

 

[mossfan563]

You need to accelerate it for 408m at 1.54 m/s², and then decelerate it for the other 408m.

So the maximum speed would be the point between accelration and deceleration. So 408 would be the correct displacement instead of 816.
Speed would be 35.449 m/s.
Does that sound right?

And for part B I would use the number I just got in the formula:
V = V_0 + at?

 

[mossfan563]

Ok I got part B wrong. How do you do part B?

 

[tiny-tim]

Ok I got part B wrong. How do you do part B?

Why do you make us guess?

Well … I'm going to guess … that you forgot to double the distance?

 

[mossfan563]

Why do you make us guess?

Well … I'm going to guess … that you forgot to double the distance?

Well I'm sorry if I'm making you guess. Trying to multitask. I know you use 816 as the displacement for this part. Acceleration is still 1.54. I'm just confused as to what formula to start with.

Maybe this formula?:
$$x = x_0 + v_0 t + (1/2) a t^2$$

 

[mossfan563]

Can anyone point me towards the right formula?