Question

1. A particle is traveling along the x-axis and its position from the origin can be modeled by x(t)=t^3 - 2t^2+ 4twhere x is centimeters and t is seconds.a. On the interval 0< t <4, find when the particle is farthest to the right.b. On the same interval, what is the maximum speed?

Answer 1

Given the relation between x and t :

where: x is the position from the origin and t is the seconds

$x(t)=t^3-2t^2+4t$

a. On the interval 0< t <4, find when the particle is farthest to the right.

There are two methods to find the answer : by graph or by derivatives

So, we will use derivatives to find the answer

$x^{\prime}(t)=3t^2-4t+4$

so, at the farthest point the speed = 0

The solution of the equation will be imaginary solutions

Which mean distance will always increase

So, the farthest point will be at t = 4

So, the farthest =

$x(4)=4^3-2\cdot4^2+4\cdot4=64-32+16=48\operatorname{cm}$

b. On the same interval, what is the maximum speed?

The speed is given by x'(t)

So, at the same interval , the speed will be maximum at t = 4

so,

$\begin{gathered} x^{\prime}(t)=3t^2-4t+4 \\ \\ x^{\prime}(4)=3\cdot4^2-4\cdot4+4=3\cdot16-16+4 \\ x^{\prime}(4)=36\text{ cm/sec} \end{gathered}$

Also, see the following figure which represents the graph solutions:

The position equation is the blue curve

The speed equation is the violate curve

The line x = 4 is the time at t = 4 seconds

The intersect with the curves give the answers