so much easier if you leave units in km and hrs.
... time for the express to stop.
... in hr , the freight train moves km
The Question I have for a Physics problem is:
A freight train is rolling down the track at 65.0km/hr. Out of the Fog, One Kilometer behind, a fast express going 120km/hr, appears on the same track. The express engineer slams on the brakes. With the brakes set, he needs 3.00km to stop, Will the express rear-end the freight train or will the express train be able to AVOID a crash?
So here is what I have done so far:
V1= 65km/hr / 3.6 = 18.1m/s
V2= 120km/hr / 3.6 = 33.3m/s
d= 3km x 1000 = 3000m
So I decided to try and find acceleration.
D = V2^2 - V1^2 / 2A
So I rearranged for Acceleration:
(2A) D = V2^2 - V1^2 / 2A (2A)
2AD = V2^2 - V1^2
2AD / 2D = V2^2 - V1^2 / 2D
A = V2^2 - V1^2 / 2D
Now to Plug the numbers in:
A = 33.3^2 - 18.1^2 / 2(3000)
A = 1108.89 - 327.61 / 6000
A= 781.28 / 6000
Am I on the right track?
Skeeter, I am not firmiliar with the formulas you are using. I've only been in Physics 11 course for 1 month, so I don't understand the formulas you are using. My understanding of physics in general is limited and I've only been given a few formulas. Also, the Reason I am going through steps is because I am to show FULL WORK. So that's why I was asking for particular steps right. It's for a project which I have to show all the steps, equations and everything, so I'd like to take it in particular steps, to check up on my work.
D= V1T + 1/2AT^2
D= (V1 + V2 /2) T
D= V2^2 - V1^2 / 2A
A= V2 - V1 / T
Here is my new attempt which matches the answer that skeeter used, only using a diffrent formula.
(For the Express Train)
A= 120^2 - 0^2 / 2(3)
14400 / 6 = 2400km/hr^2
120 / 2400 = 0.05 hour
= 1/20 hour or 3 minutes to stop
d= 65 X 1/20
Does this work?
(In this case the minimum seperation is positive but it has to be demonstrated)
let and be the respective positions in km of the freight and express trains, relative to the express train's position at t = 0 (in hrs) when it emerges from the fog.
the express train's acceleration is
the position functions for each train ...
the distance between their positions at any time t (in hrs) ...
the minimum distance occurs at hrs
at that time, km
graphs of the respective position functions are attached.