Time it would take to increase speed to..?

Question

A freight train has a mass of 3E7 kg.
If the locomotive can exert a constant pull of 7.6E5 N, how long would it take to increase the speed of the train from rest to 289800 km/s? Answer must be in units of s.

Answer 1

This is only newtonian physics calculations, which does not really apply to speeds that high but here it goes;


First, make sure all units are in kg, s, meters and N.

F = MA. F 7.6E5. and you know Mass. Calculate A.

A = something.

Then use this equation: V = V(initial) + AT
T = times and Vinitial is zero. You know A and V is final velocity.

I would imagine a really long time because 289800km/s is realllly fast...

Answer 2

You cannot ignore relativistic effects, since the train's final velocity is .967c. Relativistic mass will be m0/√(1-v^2/c^2). The accleration is then (F/m0)√(1-v^2/c^2). The velocity is ∫a*dt = ∫(F/m0)*√(1-v^2/c^2)dt. vf = ∫(F/m0)*√(1-vf^2/c^2)dt. Take the derivative of both sides to get

d(vf)/dt = (F/m0)*√(1-vf^2/c^2)

d(vf)/√(1-vf^2/c^2) = (F/m0)dt

∫1/√(1-vf^2/c^2)d(vf) = (F/m0)*t

∫[c/√(c^2-vf^2)]d(vf) = (F/m0)*t

The integral on the left evaluates to c*arcsin(v/c) so

t = (m0/F)*c*arcsin(v/c). Using c = 299792458 m/sec

this comes out 1.552*10^10 sec or 491.81 years.

Surprisingly this is not much different from the non-relativistic computation, t = vf*m0/F which gives 1.133*10^10 sec or 362.5 years

Answer 3

(2.898E8 * 3E7) / 7.6E5 sec = 1.179E10 sec

I have ignored relativistic effects, which would be considerable in this case.