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February 28, 2015

February 28, 2015

Posted by **Charlie** on Sunday, February 24, 2008 at 2:54pm.

- Physics -
**Damon**, Sunday, February 24, 2008 at 5:34pmForce from engine = 1.4*10^5 - 3.9 t

The cars do not move apart, so can be considered one rigid body twice as massive. m = 1.6 * 10^4 kg

Slope = 24 deg so component of weight down track = m g sin 24 = 1.6*10^4 * 9.8 * .407

= 6.38 *10^4 N down track

F = m a

up track

F = 1.4*10^5 - 3.9 t -6.38*10^4 = (1.6*10^4) a

so

1.6 * 10^4 a = 7.62 * 10^4 -3.9 t

a = d^2x/dt^2 = 4.76 - 2.44*10^-4 t

now if it were constant acceleration, I would assume that v = Vo + a t

so I will try something like that to solve this differential equation

try

v = Vo + B t + C t^2

then a = dv/dt = B + 2C t

well that is kind of encouraging

I will use B = 4.76 and 2C = -2.44 *10^-4

so

v = 2.9 + 4.76 t - 1.22 *10^-4 t^2

the train stops when v = 0

0 = 2.9 + 4.76 t - 1.22 *10^-4 t^2

solve that quadratic equation for t

- Physics -
**Damon**, Sunday, February 24, 2008 at 5:47pmCheck my arithmetic (and your given decrease of 3.9 N/second) because I get about 3.9*10^4 seconds which is around 10 hours.

I think you have a typo, throttle should be backed off like 3.9*10^4 N/s or something. It takes forever to back off from 1.4*10^5 N at only 3.9 N/s

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