A train is traveling up a 4.6° incline at a speed of 3.00 m/s when the last car breaks free and begins to coast without friction.
How long does it take for the last car to come to rest momentarily?
How far did the last car travel before momentarily coming to rest?
2006-09-13 04:21:28 · 3 answers · asked by otsatsa3 2 in Science & Mathematics ➔ Physics
The force of gravity acting against the motion of travel has magnitude sin(4.6) * 9.8= .79 m/s^2. Since the train is rolling at 3 m/s, the stopping time is 3/.79=3.8 seconds.
The answer to the second question is 3*3.8-1/2*.79*(3.8)^2=5.7 meters.
2006-09-13 04:35:03 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
4.6"incline implies tangent of angle=4.6
accelaration= -g sin A
where angle of incline=A
therefore time taken= (initial vel.)/ (accelaration)
and distance travelled= (initial vel.)^2/(2*(accelaration))
on feeding numerical data we get
time=0.313 second
distance=0.469 meter
2006-09-13 11:32:58 · answer #2 · answered by Punditji 1 · 0⤊ 0⤋
What's the mass of the last car????
2006-09-13 11:24:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋