suppose the position of an object moving horizontally after t seconds is given by the following function s =…

suppose the position of an object moving horizontally after t seconds is given by the following function s = f(t), where s is measured in feet, with s>0 corresponding to positions right of the origin. f(t)=t² - 10t, 0≤t≤11. a. graph the position function. b. find and graph the velocity function. when is the object stationary, moving to the right, and moving to the left? c. determine the velocity and acceleration of the object at t = 1. d. determine the acceleration of the object when its velocity is zero. e. on what intervals is the speed increasing? a. graph the position function. choose the correct graph.

suppose the position of an object moving horizontally after t seconds is given by the following function s = f(t), where s is measured in feet, with s>0 corresponding to positions right of the origin. f(t)=t² - 10t, 0≤t≤11. a. graph the position function. b. find and graph the velocity function. when is the object stationary, moving to the right, and moving to the left? c. determine the velocity and acceleration of the object at t = 1. d. determine the acceleration of the object when its velocity is zero. e. on what intervals is the speed increasing? a. graph the position function. choose the correct graph.

Answer

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position function $s(t)=t^{2}-10t$. Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=2t - 10$.

Step2: Find the acceleration function

The acceleration function $a(t)$ is the derivative of the velocity function. So, $a(t)=v^\prime(t)=\frac{d}{dt}(2t - 10)=2$.

Step3: Find when the object is stationary

The object is stationary when $v(t)=0$. Set $2t - 10 = 0$. Solving for $t$ gives $2t=10$, so $t = 5$.

Step4: Determine direction of motion

When $t<5$, $v(t)=2t - 10<0$, so the object is moving to the left. When $t>5$, $v(t)=2t - 10>0$, so the object is moving to the right.

Step5: Analyze acceleration when $v = 0$

When $t = 5$ (where $v = 0$), $a(t)=2>0$.

Step6: Analyze when speed is increasing

Speed is increasing when $v(t)$ and $a(t)$ have the same sign. Since $a(t)=2>0$ for all $t$, speed is increasing when $v(t)>0$, i.e., when $t>5$.

Step7: Graph the position function

The position function $s(t)=t^{2}-10t$ is a parabola. The vertex of the parabola $y = ax^{2}+bx + c$ is at $t=-\frac{b}{2a}$. For $s(t)=t^{2}-10t$ ($a = 1$, $b=-10$), the vertex is at $t = 5$, and $s(5)=5^{2}-10\times5=25 - 50=-25$. When $t = 0$, $s(0)=0$.

The position function $s(t)=t^{2}-10t$ is a parabola opening upwards. We can find some points: When $t = 0$, $s(0)=0$; when $t = 1$, $s(1)=1 - 10=-9$; when $t = 10$, $s(10)=100 - 100 = 0$; when $t=11$, $s(11)=121-110 = 11$.

a. To graph the position function $s(t)=t^{2}-10t$, we know it is a parabola $y=x^{2}-10x$ with vertex at $(5, - 25)$ and $y -$intercept at $(0,0)$. b. $v(t)=2t - 10$, graph of $y = 2x - 10$ is a straight - line with slope $2$ and $y -$intercept $-10$. The object is stationary at $t = 5$ (where $v = 0$), moving left for $t<5$ ($v<0$) and moving right for $t>5$ ($v>0$). c. $a(t)=2$, it is a horizontal line. When $t = 1$, $a(1)=2$. d. The object is stationary at $t = 5$. It is moving to the left when $t\in[0,5)$ and moving to the right when $t\in(5,11]$. e. Since $a(t)=2>0$ and speed is increasing when $v(t)>0$, speed is increasing on the interval $(5,11]$.

Answer:

a. The graph of $s(t)=t^{2}-10t$ is a parabola opening upwards with vertex at $(5,-25)$ and passing through $(0,0)$ and $(10,0)$. b. $v(t)=2t - 10$, object is stationary at $t = 5$, moves left for $t<5$ and right for $t>5$. c. $a(t)=2$, $a(1)=2$. d. Stationary at $t = 5$, left on $[0,5)$, right on $(5,11]$. e. Speed is increasing on $(5,11]$.