Calculating Braking Distance for Trains: 72 km/h and 133 km/h Speeds

[somnium]

I guess my brain is fried, I can't figure out what formula to use for this problem:

Two trains, one traveling at 72 km/h and the other at 133 km/h, are headed toward one another along a straight, level track. When they are 860 m apart, each engineer sees the other's train and applies the brakes. The brakes decelerate each train at the rate of 1.0 m/s2.
What is the braking distance for each train?

 

[Valhalla]

kinematics...the are both accelerating (I hate the word decelerate) opposite of their velocity at a constant rate.

Can you think of any kinematics equation that relates initial velocity, final velocity, acceleration, and distance?

 

[somnium]

x = x0 + v0t + (1/2)at2

?

 

[Cyrus]

That will work, but how will you go about solving two variables, t, and x?

 

[somnium]

i just got that one, but here is another:

To stop a car, you require first a certain reaction time to begin braking. Then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.0 km/h, and 24.4 m when the initial speed is 47.8 km/h.

What is your reaction time?
What is the magnitude of the deceleration?

I think I'm completely missing the point in this one...

 

[Cyrus]

This my be useful, the amount of reaction time will be the same in both cases, and the reaction time will be incorperated in the total time it takes to stop the total distance. Part of the distance you will move 80km/hr and 47.8km/hr from t_inital to t_reaction amount of time. Then you will have constant deceleration from t_reaction to t_stop.

 

[somnium]

i still don't quite get how i am to plug this into a formula.
i'm assuming that the wording means that the car starts out at 80km/h,
displaces 56.7m, then the velocity changes to 47.8km/h and the car moves another 24.4m?
if this is so, does reaction time mean the time it takes to change velocities?

 

[geoorge]

i got it just 5 minutes to get done

 

[Cyrus]

You will have to think on your own for this one. Thats what you get for just pluging and chuging and not taking the time to read the theory of the formulas.


No, the wording states two different cases,

case 1,

56.7 m when its initial speed is 80.0 km/h

case 2,

24.4 m when the initial speed is 47.8 km/h.

I have noticed that you have posted a lot of homework questions tonight. This should be for homework help, try to do it on your own. If you don't know what formulas you should use, it would help you to go back and read the material one more time to understand what's going on.

 

[somnium]

You will have to think on your own for this one. Thats what you get for just pluging and chuging and not taking the time to read the theory of the formulas.

i'm sorry if it appears that way to you, but no, i am not just plugging and chugging. this is all material that i learned a long time ago, and now must remember the formulas to. the textbook i am using is not exactly integrated with the online questions that i am answering.

and thus, i am asking for hints from those who recall this stuff quicker than i am, such as the one i got from Valhalla, who simply gave me a link to a very useful site with the necessary formulas. (THANKS, VALHALLA!)

 

[somnium]

and is asking for guidance in solving 3 problems really so excessive?

 

[geoorge]

V1^2 + 2aX1 = 0 (VF)
V2^2 + 2aX2 = 0 (VF)

THEN

V1^2 + 2aX1 =V2^2 + 2aX2

LETS SAY

X1=860MTS-X2

V1^2 + 2a(860-X2) =V2^2 + 2aX2 DOING THE MATH WE HAVE

(V1^2 - V2^2 )/4a + 430 = X2

V1=20m/s
V2=37m/s

remeber the a must be minus

then we have X2=671.25mts and X1 =860-672.25 =188.75mts

but the trains will crash because their remainig velocity will be

VF1=4.99
VF2=4.99

 

[geoorge]

but the second train running at 20m/s will only need 200 mts to brake, so it would be like 20 seconds , mean while the other train will need 36 seconds to get to the fatal 600mts, so the people of one of the trains has at least 16 seconds to leave!