Question

1. (3 points) the table gives the position of a particle moving along the x - axis as a function of time in seconds, where x is in meters. give an estimate of the instantaneous velocity of the object at time t = 4 seconds. use the average of left - hand and right - hand approximations. give units.

t	0	2	4	6	8
x(t)	0	14	-6	-18	-4

Explanation:

Step1: Calculate left - hand approximation

The left - hand approximation of the derivative (velocity) at $t = 4$ uses the points at $t=2$ and $t = 4$. The formula for the average rate of change (average velocity) is $\frac{\Delta x}{\Delta t}=\frac{x(t_2)-x(t_1)}{t_2 - t_1}$. Here, $t_1 = 2$, $t_2=4$, $x(2)=14$ and $x(4)=-6$. So, $v_{left}=\frac{x(4)-x(2)}{4 - 2}=\frac{-6 - 14}{2}=\frac{-20}{2}=-10$ m/s.

Step2: Calculate right - hand approximation

The right - hand approximation of the derivative at $t = 4$ uses the points at $t = 4$ and $t=6$. Using the formula $\frac{\Delta x}{\Delta t}=\frac{x(t_2)-x(t_1)}{t_2 - t_1}$, with $t_1 = 4$, $t_2 = 6$, $x(4)=-6$ and $x(6)=-18$. So, $v_{right}=\frac{x(6)-x(4)}{6 - 4}=\frac{-18+6}{2}=\frac{-12}{2}=-6$ m/s.

Step3: Calculate the average of left - hand and right - hand approximations

The instantaneous velocity estimate $v$ is the average of $v_{left}$ and $v_{right}$. So, $v=\frac{v_{left}+v_{right}}{2}=\frac{-10+( - 6)}{2}=\frac{-10 - 6}{2}=\frac{-16}{2}=-8$ m/s.

Answer:

$-8$ m/s