http://prntscr.com/16cd1n6 I need help with part B only for this question

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11) If v is the velocity, v=t(1-t2)=t-t3.

acceleration=dv/dt=t-3t2=0 when acceleration is maximum (or minimum).

Solving for t, t2=⅓, so t=1/√3.

d2v/dt2=1-6t. When t=1/√3, d2v/dt2<0, so the velocity is maximum at t=1/√3, and it has the value:

v=(1/√3)(1-⅓)=2/(3√3)m/s or m/s-1, which can also be written 2√3/9m/s.

10) a) Driving force of the engine is 10000N, less the resistance to motion of the engine and carriage (5000+3000=8000N)=2000N. Using Newton's Law F=ma, a=F/m=2000/18000, where m is the combined mass of the engine and carriage=12000+6000=18000kg, so acceleration a=1/9ms-2, which I think you knew.

b) When the shunt fails (detaches), the resistance of the carriage alone is the slowing force which will bring the carriage to a halt. So applying F=ma to the carriage: m=6000kg, F=-3000N, so a=-3000/6000=-½ms-2.

v=u+at=0 where u is the initial velocity=30ms-1. The final velocity is of course zero.

30-t/2=0, t=60 seconds to come to a stop.

c) (which I think you know) Any contribution to the energy by the shunt is ignored, including its effect on the velocity of the carriage when it fails.

by Top Rated User (884k points)