Acceleration 7124

The train passes 700m, braking with an acceleration of -0.15 m/s2. How long does it break, and what is the final speed of the train if the initial was 55 km/h?

Correct answer:

t =  69.5956 s
v2 =  4.8384 m/s

Step-by-step explanation:

s=700 m a=0.15 m/s2 v1=55 km/h m/s=55:3.6  m/s=15.27778 m/s  s=v1 t21 a t2 700=15.277777777t0.50.15t2  700=15.277777777 t0.5 0.15 t2 0.075t215.278t+700=0  a=0.075;b=15.278;c=700 D=b24ac=15.278240.075700=23.4104938034 D>0  t1,2=2ab±D=0.1515.28±23.41 t1,2=101.85185185±32.256261272225 t1=134.10811311889 t2=69.595590574442   Factored form of the equation:  0.075(t134.10811311889)(t69.595590574442)=0  t>0 v2>0 t=t2=69.5956=69.5956 s

Our quadratic equation calculator calculates it.

v2=v1a t=15.27780.15 69.5956=4.8384 m/s



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