The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typicallyµ,,= 0.002. suppose a 180,000 kg locomotive is rolling at 10 mis (just over 20 mph) on level rails. if the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? how far will the locomotive move during this time?

Explanation:

Given

coefficient of kinetic friction

mass of locomotive=180,000 kg

v=10 m/s

deceleration provided by friction

Time taken to stop completely is

v=u+at

where v=final velocity

u=initial velocity

a=acceleration

t=510.20 s

Distance traveled during this point

s=2551.02 m

carla slowed down in the beginning of her trip, stopped, and then sped up

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.