Post a New Question

Physics

posted by .

A train travels 84 km/h when it reaches a slower train 47 meters ahead traveling the same direction at 24 km/h. If the faster train decelerates at 2.1 m/s^2 while the slower train is at a constant speed, when will they collide? If the train that is faster begins to decelerate at 2.1 m/s^2 while the slower train continues at a constant speed, what willbe the relative speed in which they will collide?

I got 4.9 seconds for the first one and I do not get the last one. Thanks Y'all.

  • Physics -

    84 km/hr *(1000m/km) /(3600 s/hr)= 23.3 m/s
    24 km/hr = 6.67 m/s

    d fast train = d
    d slow train = d-47

    d fast train= d = 23.3 t - .5(2.1) t^2
    d slow train = (d-47) = 6.67 t
    so
    23.3 t - 1.05 t^2 = 6.67 t + 47
    1.05 t^2 - 16.6 t + 47 = 0
    approximate
    2t = 26.6 +/- sqrt (276-188)
    2 t = 26.6 +/- 9.4
    2t = 36 or 17.2
    t = 18 or 8.8
    the 8.8 is the first overtaking, the 18 is when the slow one then catches up with the fast one again after it really slows down (assuming they are actually on parallel tracks and do not crash)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question