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 the each train what would I have to do?
To find the speed of each train at impact, you can use the equation of motion:
v^2 = u^2 + 2ad
Where:
v is the final velocity (which is 0 in this case since the trains come to a stop),
u is the initial velocity (the speed of the train),
a is the acceleration (deceleration in this case, which is -1 m/s^2), and
d is the distance (800m for the faster train).
So, the equation becomes:
0 = u^2 + 2(-1)(800)
Simplifying the equation:
0 = u^2 - 1600
Rearrange the equation to isolate u^2:
u^2 = 1600
Taking the square root of both sides:
u = ±40
Since the speed of a train cannot be negative, the speed of the faster train is 40 m/s.
Now, to find the speed of the slower train, we can use the relationship given in the hint: the distance the slower train travels is half the distance the faster train travels.
So, the distance traveled by the slower train is (1/2) * 800 = 400m.
Using the same equation of motion:
0 = u^2 + 2(-1)(400)
Simplifying the equation:
0 = u^2 - 800
Rearrange the equation to isolate u^2:
u^2 = 800
Taking the square root of both sides:
u = ±28.3
Since the speed of a train cannot be negative, the speed of the slower train is approximately 28.3 m/s.
Therefore, at impact, the speed of the faster train is 40 m/s and the speed of the slower train is approximately 28.3 m/s.