The motion of a set of particles moving along the x-axis is governed by the differential equation dx/dt=t^3-x^3 where x(t) denotes the position at time t of the particle. a. if a particle is located at x=1 when t=2, what is its velocity at this time? b. show that the acceleration of a particle is given by d^2x/dt^2=3t^2-3t^3x^2+3x^5 c. if a particle is located at x=2 when t=2.5, can it reach the location x=1 at any later time? (hint: t^3-x^3=(t-x)(t^2+xt+x^2)

gives the velocity at any point , so you need only plug these values in:



Acceleration can be determined by differentiating the velocity function:







In order for any particle to reach the point , we require that be decreasing for some value of . In other words, must be negative for some . This basically amounts to solving the inequality,



Note that the left hand side is negative when either or , but not both.

When , in the first case you have . But this isn't possible, since (time) must always be growing in the positive direction.

Meanwhile, in the second case, you have



for all , which means can never be negative, which in turn means must be increasing. Therefore it's not possible for any particle to reach a position of for any .

Dis a correct answer