Asked by Bob
The engineer of a passenger train moving at 29.0 m/s sees a freight train 360.m ahead on the track moving at 6.00 m/s. The engineer takes 0.400 seconds to react, and then applies the brakes. At what rate must the passenger train then decelerate in order to avoid a collision?
The answer I got was -0.703 m/s^2 but the answer is actually -0.754 m/s^2.
The answer I got was -0.703 m/s^2 but the answer is actually -0.754 m/s^2.
Answers
Answered by
Damon
freight train goes 6 t
our train goes 6t+360
Vi = 29
Vf = 6
a assumed negative
v = Vi - a t
6 = 29 - a t
a t = 23
d = 6t+360 = 29t-(1/2)a t^2
360 = 23 t -(1/2)(23 t)
720 = 23 t
t = 31.3 seconds to crash
a = .74
whoops, forgot the .4 seconds to react
we go 14.5 m
he goes 2.4 m
so real distance between is a little smaller
360 - 12.5 = 347.5
347.5 = 23 t -(1/2)(23 t)
694 = 23 t
t = 30.17 seconds
a * 30.17 =23
a = .762
our train goes 6t+360
Vi = 29
Vf = 6
a assumed negative
v = Vi - a t
6 = 29 - a t
a t = 23
d = 6t+360 = 29t-(1/2)a t^2
360 = 23 t -(1/2)(23 t)
720 = 23 t
t = 31.3 seconds to crash
a = .74
whoops, forgot the .4 seconds to react
we go 14.5 m
he goes 2.4 m
so real distance between is a little smaller
360 - 12.5 = 347.5
347.5 = 23 t -(1/2)(23 t)
694 = 23 t
t = 30.17 seconds
a * 30.17 =23
a = .762
Answered by
Damon
360 - 12 = 348
348 = 23 t -23t/2
696 = 23 t
t = 30.3
a = 23/30.3 = .759 m/s^2
348 = 23 t -23t/2
696 = 23 t
t = 30.3
a = 23/30.3 = .759 m/s^2
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