The position of a particle moving along the x axis depends on the time according to the equation x = ct3 - bt6, where x is in meters and t in seconds. let c and b have numerical values 2.9 m/s3 and 2.0 m/s6, respectively. from t = 0.0 s to t = 1.3 s, (a) what is the displacement of the particle? find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.

Given the equation of the particle, we know that:

a) In we evaluate in the former equation:

In

a) So, we know that the displacement of te particle is given by:

To find it's velocity, we need to derivate the equation of position by the formula:

And evaluate this expression at each specified t:

b)

c)

d)

e)

To find it's acceleration, we need to derivate the equation of velocity by the formula:

And evaluate this expression at each specified t:

f)

g)

h)

i)

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.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y