Expert: Alon Mandes - 2/12/2010

Question
my teacher just went on maternity leave as she left us with a set of questions, this is one of them an i have no idea what to do.

A particle moves along the x-axis so that at any time t>0 its velocity is given by v(t)= t ln t-t.  At time t=1, the position of the particle is x(1)=6.

a. write an expression for the acceleration of the particle.
b. for what values of t is the particle moving to the right?
c. what is the miniumum velocity of the particle?
d. write an expression of the position x(t) of the particle.

Answer
As we know, if x(t) is the function expressing the position of a particle, then :
1. v(t)=dx(t)/d(t)=x'(t) is the function expressing the velocity of a particle .
2. a(t)=dv(t)/dt=v'(t)=x''(t) is the function expressing the acceleration of a particle.

Let's answer our problem :
a. a(t)=v'(t)=[t*Ln(t)-t]'=Ln(t)+t*(1/t)-1=Ln(t) .
b. the particle is moving to the right when v(t)>0 : t*Ln(t)-t>0 -->Ln(t)>1 --> t>e .
c. Minimum velocity occurs when the acceleration is zero, thus, Ln(t)=0 : t=1 .
  v(1)=1*Ln(1)-1=-1
d. x(t)=∫v(t)dt+C=∫tLn(t)dt-∫tdt+C=(1/4)t²[2Ln(t)-1]-(t²/2)+C . To find C we use the initial
  condition : x(1)=6 --> x(1)=(1/4)(-1)-(1/2)+C=6 --> C-(3/4)=6 --> C=27/4 .

Alon.