A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the

Question

3 years ago

7

train move away from the track gang and what was its speed after the 5 min.?

1 answer:

Amanda [17] · Answer 1 · 2022-07-19T00:42:21+03:00

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check: V0 = 5 m/s

V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s the speed after 300 sec