Answered

At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.