Question

http://prntscr.com/16ccxmz Please help.

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

Answer 1

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.