# A red train traveling at 72 km/h and a green train traveling at 144 km/h are headed toward one another along a straight, level track. When they are 950m apart, each engineer sees the other’s train and applies the brakes. The brakes decelerate each train at the rate of 1 m/s^2. Is there a collision? If so what is the speed of each train at impact? If not what is the separation between the trains when they stop?

I will be happy to critique this for you. Hint: The distance the slower train travels is 1/2 the faster train. So see if the slow train can stop in 950(2/3) m.

Vf^2= vo^2 - 2ad
change Vo to m/s If you get a solution here with vf=0 for d less than 2/3 950, then they do not collide.

I got d as 800m for the faster train. so they do collide. To find the speed of each train what would I have to do?

We have two objects and velocities: red 72km/h, green 144km/h.
You should convert each velocity to m/s for compatibility. 1km/hr = 1000m/3600s = 10m/36s =5m/18s.
(BTW, you should learn some basic conversions. It's definitely worth memorizing this one.)
The velocity of the red train is
72*5/18 m/s= 20m/s and the green train's velocity is 40m/s
Let's look at the distance the red one travels in decelerating. The formula is given by
(1)v_f^2= v_o^2 + 2ad so d=(v_f^2 - v_o^2)/2a
For the red train v_f=0m/s, v_o=20m/s
and a=-1m/s^2 then d=(0-400)/-1 m = 400m
For the green train v_f=0m/s, v_o=40m/s and a=-1m/s^2 then
d=(0-1600)/-1 m = 1600m
They do collide.
I should point out this is not a linear equation, so we can't simply double the distance for the second one.
The distance needed for each is proportional to the square of the velocity.
Let d1 be the distance the red one travels and d2 the distance for the green one. The distances are related by
d1+d2=950m and d1:400=d2:1600 so d2=4d1 and
5d1=950 or d1=190 and d2=760
Now using (1) above for the red train, we have
v_f=sqrt(v_o^2 + 2ad) where v_o=20m/s and a=-1m/s^2 so v_f= sqrt(20)m/s
For the green train the calculations are similar.
v_f=sqrt(v_o^2 + 2ad) where v_o=40m/s and a=-1m/s^2 so v_f= sqrt(1220)m/s
In this problem it appears the green train barely starts to decelerate when they collide.

Be sure to check my calculations and that I used the formulas correctly.
I only worked this because it's been asked about 5 times now and you'll probably need this for either a quiz or test soon.

I see I got the velocity v_f of the green train wrong. I used the same distance for it as the red one. The formula is
v_f=sqrt(v_o^2 + 2ad) where v_o=40m/s and a=-1m/s^2 and d=760m so
v_f=sqrt(1600-2*(-1)*760)=sqrt(80) for the green train.
I 'think' I got the red one right, but check it just the same.

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1. GOOD!!!thanks you

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2. (1)v_f^2= v_o^2 + 2ad so d=(v_f^2 - v_o^2)/2a
For the red train v_f=0m/s, v_o=20m/s
and a=-1m/s^2 then d=(0-400)/-1 m = 400m

this statement is wrong-- you failed to multiply the acceleration by 2 as it states in the equation. So (vx^2-vx0^2)/2(ax)= x
so therefore--(0-400)/(2*-1)= 200m=x

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3. there also is a problem with the vaster train's distance. Make sure to divide by two. Dividing in this instance makes the d of the faster train 800 instead of 1600. These different numbers don't change the fact that they collided, though, considering d1+d2=total, which is 200+800=1000, which is greater than 950.

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4. So I'm still confused on what's happening here?

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