THe French National Railroad holds the world's speed record for passenger trains in regular service. A TGV train traveling at a speed of 3400 km/r requires 1.20 km to come to an emergency stop. Find the braking acceleration for this train, assuming constant acceleration.
2006-08-26 11:32:56 · 9 answers · asked by Anna 1 in Science & Mathematics ➔ Physics
opppsss sorry the trains traveling 340 km/hr
2006-08-26 16:03:06 · update #1
Vf = final velocity, Vi = initial, a = acceleration, d= distance
Vf^2 = Vi^2 + 2ad
0 = (3400km/hr)^2 + 2a(1.2km)
-(3400km/hr)^2 = (2.4 km)a
Divide through, and you get...
-4816666.6km/hr^2? Sounds like a huge value. Are you sure the train is traveling at 3400 km/hr? If it is, then that should be the correct answer...
If you would like more help or have any other questions, then please e-mail or contact me in someway, I'll be willing to help.
2006-08-26 11:42:40 · answer #1 · answered by Link 5 · 0⤊ 0⤋
First off 3400 km/hr is about 3 times the speed of sound. The land speed record of a manned vehicle is 1227 km/hr. So, your example is not correct.
Assuming that the deceleration is constant, the mean speed during deceleration is 1700 km/hr. It would take 1.2/1700 hrs to decelerate, or 2.54 seconds. 3400/2.54 would give you a constant deceleration rate of about 1338 km/sec. Quite enough to kill everyone on board.
I'd check your problem before answering it.
2006-08-26 19:12:31 · answer #2 · answered by SPLATT 7 · 1⤊ 0⤋
the formula for constant acceleration is
(Vf)^2 - (Vi)^2 = 2ad
where Vf is the final velocity which in this case is 0 (the car stops)
Vi is the intial velocity = 3400 km/...
a is the acdelration
d is the disatnce = 1.20 km
and find the a.
2006-08-26 18:44:17 · answer #3 · answered by ___ 4 · 0⤊ 0⤋
ok....using 340 km/hr
v^2(final) = v^2(initial) + 2as
final velocity is 0 therefore....
0 = (340)^2 + 2a(1.2)
solving for a
a = -48166.67 km/hr^2 (negative cause it's slowing down)
2006-08-26 18:42:25 · answer #4 · answered by mdc 2 · 0⤊ 0⤋
To solve this question, use the equation:
v^2 = u^2 + 2as
where
v = final velocity (m) [zero in this case!]
u = initial velocity (ms^-1) 340000/3.6 in this case]
s= distance 1200 m in this case
u = 340/3.6 ms^-1
I hope that helps. If you need more help, please re post or email me.
2006-08-26 18:43:11 · answer #5 · answered by Auriga 5 · 0⤊ 0⤋
its 4.82*10^6 km/hr^2
2006-08-26 23:20:40 · answer #6 · answered by avik r 2 · 0⤊ 0⤋
133.79 km/s^2
2006-08-26 18:53:07 · answer #7 · answered by Vik4u 2 · 0⤊ 0⤋
one thing, it braking decelleration not acceleration
2006-08-26 21:17:59 · answer #8 · answered by {««мα∂gυу»»} 2 · 0⤊ 1⤋
sorry. i can't help. good question tho :)
2006-08-26 18:36:05 · answer #9 · answered by bogo_648 1 · 0⤊ 1⤋