Answer to Question #341051 in Calculus for Mutamba

Question

Consider a particle moving along the x-axis where x(t) is the position of the

particle at time t, x′(t) is its velocity, and x

′′(t) is its acceleration.

If x(t) = t3

− 6t2 + 9t − 2, 0 ≤ t ≤ 5

(a) Find the velocity and acceleration of the particle. (b) Find the open

t-

intervals on which the particle is moving to the right. (c) Find the velocity of

the particle when the acceleration is 0

Answer 1

$v(t)=x'(t)=3t^2-12t+9$

$a(t)=x''(t)=6t-12$

b) For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t² -12t +9

We can factorize this function as v(t)= 3 (t²- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0≤t≤10 we need to analyze the following intervals:

0 < t < 1 ; 3 < t < 10

c)$a(t)=0=6t-12$

t=2

$v(2)=3(2^2)-12(2)+9=-3$