A trains departs from a station and accelerates at a constant rate of 32 feet per second per second. What is the speed of the train after one hour? If the train encouncters a curve, is it likely to derail?
2007-12-03 18:22:11 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Or is the train more likely to burn up before encountering a curve?
2007-12-03 18:24:14 · update #1
One hour = 3600 seconds, so the speed of the train is 32*3600 = 115200 feet per second = 78545 mph. I would say that it is very likely to derail, and most likely something will burn out before it reaches that speed.
2007-12-03 18:48:59 · answer #1 · answered by gp4rts 7 · 1⤊ 0⤋
So this is how Santa gets around!!! I think his trajectory was about 3,600 miles per second, but it could be argued that calculation is in orbit.
So if you are talking about a train in space without atmospheric friction, it would be much different than a train
on the earth, which would most likely burn with friction at the speeds that you are suggesting.
I'd go with Santa. He probably has some special tiles that allow re-entry into the atmosphere without burning up all our presents. Happy Holidays, everybody...
Ken
2007-12-03 23:46:07 · answer #2 · answered by Ken C 3 · 1⤊ 0⤋
The equation you want to apply is v_f=v_o+at, the position v_f is very last speed, v_o is initial speed, a is acceleration, and t is time. i imagine what the priority implies is that the practice stopped on the station, which signifies that its speed change into 0 on the station. The practice reaches 4.8m/s in 5.3s, so that you'll discover the acceleration: 4.8 m/s=0m/s + a*5.3s. detect a. Now that you've a, you may settle on the priority. the recent initial speed is 4.8. The time elapsed is 7.0s-5.3s. The acceleration remains a similar. Plug the values in: v_f=4.8m/s+a*(7.0s-5.3s) settle on for v_f and also you've were given your answer. :)
2016-10-25 10:18:24 · answer #3 · answered by hodnett 4 · 0⤊ 0⤋
i think the speed of the train after 1 hour is going to be..
as v = u + at
v = 0 + 32*3600 = 115200 ft/sec........
hmmmm i think this speed is 'quite' fast than a average bullet train.... the speed of train will be 35112.96 m/s...... or simply 21.81818182 miles/sec....
i will fly away in the air creating a sonic boom
2007-12-03 19:41:34 · answer #4 · answered by Anonymous · 1⤊ 0⤋