Answer to Question #341051 in Calculus for Mutamba

Question #341051

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


1
Expert's answer
2022-05-17T08:58:00-0400

"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"


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