Asked by A
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?
Answers
Answered by
Damon
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 !
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 !
Answered by
Axel
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.
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