1. Physics- Kinematics (Grade 11 Level)

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:

Variable List:

V1= 65km/hr / 3.6 = 18.1m/s
V2= 120km/hr / 3.6 = 33.3m/s
a= ?
d= 3km x 1000 = 3000m
t= ?

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 / 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

A= 0.13m/s^2

Am I on the right track?

2. so much easier if you leave units in km and hrs.

$
\Delta x = \frac{1}{2}(v_0 + v_f)t
$

$
3 = \frac{1}{2}(120 + 0)t
$

$t = \frac{1}{20} \, hr$ ... time for the express to stop.

... in $\frac{1}{20}$ hr , the freight train moves $65 \cdot \frac{1}{20} = 3.25$ km

3. 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.

These:

D= 1/2GT^2

D= V1T + 1/2AT^2

D= (V1 + V2 /2) T

D= V2^2 - V1^2 / 2A

A= V2 - V1 / T

4. Originally Posted by (?)G
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.

These:

D= 1/2GT^2

D= V1T + 1/2AT^2

D= (V1 + V2 /2) T I used this equation ... twice.

D= V2^2 - V1^2 / 2A

A= V2 - V1 / T
...

5. Ok... nevermind...

6. the formulae used by skeeter are taken from the geometrical properties of the values considered when they are graphed

7. What would be a good scale to graph this with using a d vs t graph?

8. Here is my new attempt which matches the answer that skeeter used, only using a diffrent formula.

(For the Express Train)

V1= 120km/hr

V2= 0km/hr

a= ?

d= 3km

t= ?

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= vt

d= 65 X 1/20

d= 3.25km

Does this work?

9. Originally Posted by skeeter
so much easier if you leave units in km and hrs.

$
\Delta x = \frac{1}{2}(v_0 + v_f)t
$

$
3 = \frac{1}{2}(120 + 0)t
$

$t = \frac{1}{20} \, hr$ ... time for the express to stop.

... in $\frac{1}{20}$ hr , the freight train moves $65 \cdot \frac{1}{20} = 3.25$ km
While the seperation between the trains has increased by the time the express has come to a stop assuming no collision in the interim, this does not in principle exclude the possibility of impact earlier. You need to show that the seperation between the trains is never zero for positive time.

(In this case the minimum seperation is positive but it has to be demonstrated)

CB

10. Originally Posted by CaptainBlack
While the seperation between the trains has increased by the time the express has come to a stop assuming no collision in the interim, this does not in principle exclude the possibility of impact earlier. You need to show that the seperation between the trains is never zero for positive time.

(In this case the minimum seperation is positive but it has to be demonstrated)

CB
the Captain makes an excellent point.

let $x_1$ and $x_2$ 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 $a = \frac{v_f^2 - v_0^2}{2\Delta x} = \frac{0 - 120^2}{6} = -2400 \, km/hr^2$

the position functions for each train ...

$x_1 = 3 + 65t$

$x_2 = 120t - 1200t^2
$

the distance between their positions at any time t (in hrs) ...

$x_1 - x_2 = 3 - 55t + 1200t^2$

the minimum distance occurs at $t = \frac{55}{2400}$ hrs

at that time, $x_1 - x_2 \approx 2.37$ km

graphs of the respective position functions are attached.