Answered

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

Sagot :

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

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