A train has a velocity of 117 km/ h and discover another train standing stil 1200 meters away. The train starts immediately to brake. After 10 seconds the other train finally starts to move away from the first train. The acceleration is equable of both trains. The first trains acceleration is -0.2 m/ s2. What must be the lowest acceleration of the second train if a collision will be avoided?
117 km/h *1000 m/km / 3600 s/h = 32.5 m/s
Train 1
x = 32.5 t -.5 *.2* t^2
x = 32.5 t - .1 t^2
Train 2
x = 1200 + .5 a (t-10)^2
when do they hit?
32.5t-.1t^2=1200+.5a(t^2-20t+100)
32.5t-.1t^2=1200+ .5at^2-10 at +50 a
(.5a+.1)t^2 -(32.5+10a)t +(50a+1200)= 0
when does t become unreal?
when b^2-4ac = 0
(32.5+10a)^2 -4(.5a+.1)(50a+1200) = 0
1056.25+650a+100a^2-100a^2-2420a-480 = 0
-1770a + 576.25 = 0
a = .326 m/s^2
check my arithmetic !
Why does the t become unreal when b^2-4ac = 0? I am trying to do this with pq-form to understand, haven't used the quadratic formula.
To find the minimum acceleration of the second train required to avoid a collision, we need to determine the distance covered by each train within the 10-second interval.
Let's start by converting the velocity of the first train from km/h to m/s:
117 km/h = (117 * 1000) / 3600 m/s = 32.5 m/s
Since the velocity is constant and the first train starts braking immediately, the distance covered by the first train in these 10 seconds can be calculated using the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given:
initial velocity (u) = 32.5 m/s
time (t) = 10 s
acceleration (a) = -0.2 m/s²
distance covered by the first train (d1) = u * t + (1/2) * a * t^2
d1 = 32.5 * 10 + (1/2) * (-0.2) * (10^2)
= 325 - 1 * 10
= 315 meters
The second train must travel a distance greater than or equal to 1200 meters to avoid a collision. Therefore, the minimum distance covered by the second train in these 10 seconds can be calculated as:
distance covered by the second train (d2) = 1200 - d1
= 1200 - 315
= 885 meters
Now, let's find the minimum acceleration of the second train needed to cover this distance within 10 seconds.
Using the same equation: distance = initial velocity * time + (1/2) * acceleration * time^2
We know:
initial velocity (u) = 0 (since the second train is initially at rest)
time (t) = 10 s
acceleration (a) = ?
d2 = u * t + (1/2) * a * t^2
Plugging in the values we know:
885 = 0 * 10 + (1/2) * a * (10^2)
885 = (1/2) * 100 * a
2 * 885 = 100a
1770 = 100a
Dividing both sides by 100:
a = 1770 / 100
a = 17.7 m/s²
Therefore, the minimum acceleration of the second train required to avoid a collision is 17.7 m/s².