Question

A particle moves along a straight line and its position at time t is given by

s(t)=2t^3−24t^2+90t t>=0 where s is measured in feet and t in seconds.
(B) Use interval notation to indicate the time interval(s) when the particle is speeding up and slowing down.


(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

Answer 1

To determine the time interval when the particle is speeding up and slowing down, find the intervals where the velocity is increasing and decreasing.

Step-by-step explanation:

The given position function of the particle is s(t) = 2t³-24t²+90t, where s is measured in feet and t in seconds. To determine the time interval when the particle is speeding up and slowing down, we need to find the intervals where the velocity is increasing and decreasing. The velocity function v(t) is the derivative of the position function with respect to time, which is v(t) = 6t² - 48t + 90.

When the velocity is positive and increasing, the particle is speeding up. The velocity is positive for values of t such that v(t) > 0. When the velocity is negative and decreasing, the particle is slowing down. The velocity is negative for values of t such that v(t) < 0.

Answer 2

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

if a(t) >0, it means 12t-48>0, and t >4, I = ]4, infinity[

the particle is speeding down

if a(t) < 0, it means 12t-48>0, and t < 4, I = ] -infinity, 4[

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

moving backward if v(t)< 0

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

so after taking care the sign of 6t² - 48 t +90,

v(t)> o when t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward

v(t) < o when t is in I = ]1, 4[,the particle is moving backward