the function describes the motion of a

Question

The function s(t) describes the motion of a particle along a line.

s(t) = t³ − 5t² + 4t

(a) Find the velocity function of the particle at any time t ≥ 0.

v(t) = ______

 

(b) Identify the time intervals on which the particle is moving in a positive direction. (Enter your answer using interval notation.)

 

(c) Identify the time intervals on which the particle is moving in a negative direction. (Enter your answer using interval notation.)

 

(d) Identify the time(s) at which the particle changes direction. (Enter your answers as a comma-separated list.)

t = ______

 

Answer 1

(a) To understand this problem, you need to know that v(t) = s'(t)

Therefore with s(t) = t³-5t²+4t

v(t)=3t²-10t+4

For (b)(c)(d)

Velocity is the change in position over the change in time. To find when the position is moving in a positive/negative direction, you must set v(t)=0 and solve for t

t=[10+-sqrt(100-48)]/6 = [10+-sqrt(52)]/6=[10+-2sqrt(13)]/6=[5+-sqrt(13)]/3

You know that v(t) will look like a parabola due to it being a quadratic equation.

With coefficient 3 being a positive number, the graph starts off being positive and then passes [5-sqrt(13)]/3 and goes negative.

It will then curve back up and pass [5+sqrt(13)]/3 and go positive again. Such is the look of a parabola.

With that in mind, whenever the graph of v(t) is positive, s(t) is moving in a positive direction and visa versa.

SO

(b) Using the explanation above,

(-inf,[5-sqrt(13)]/3 ) U ([5+sqrt(13)]/3,inf)

(c) Using the explanation above,

([5-sqrt(13)]/3,[5+sqrt(13)]/3)

(d) For all values of t when v(t)=0

Therefore, t = [5-sqrt(13)]/3,[5+sqrt(13)]/3