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.\na. graph the position function.\nb. find and graph the velocity function. when is the object stationary, moving to the right, and moving to the left?\nc. determine the velocity and acceleration of the object at t = 1.\nd. determine the acceleration of the object when its velocity is zero.\ne. on what intervals is the speed increasing?\nf(t)=24t - 4t^{2}, 0≤t≤8\na. graph the position function. choose the correct graph below

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.\na. graph the position function.\nb. find and graph the velocity function. when is the object stationary, moving to the right, and moving to the left?\nc. determine the velocity and acceleration of the object at t = 1.\nd. determine the acceleration of the object when its velocity is zero.\ne. on what intervals is the speed increasing?\nf(t)=24t - 4t^{2}, 0≤t≤8\na. graph the position function. choose the correct graph below

Answer

Explanation:

Step1: Recall the relationship between position and velocity

Velocity $v(t)$ is the derivative of position function $s(t)$. Given $s(t)=24t - 4t^{2}$, by the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=\frac{d}{dt}(24t-4t^{2})$. $v(t)=24 - 8t$

Step2: Recall the relationship between velocity and acceleration

Acceleration $a(t)$ is the derivative of velocity function $v(t)$. So $a(t)=\frac{d}{dt}(24 - 8t)$. $a(t)=- 8$

Step3: Analyze when the object is stationary

The object is stationary when $v(t)=0$. Set $24 - 8t = 0$. $8t=24$, so $t = 3$.

Step4: Analyze when the object is moving right and left

The object is moving to the right when $v(t)>0$. Solve $24 - 8t>0$, we get $8t<24$, $t < 3$. The object is moving to the left when $v(t)<0$. Solve $24 - 8t<0$, we get $8t>24$, $t>3$.

Step5: Find velocity and acceleration at $t = 1$

Substitute $t = 1$ into $v(t)$ and $a(t)$. $v(1)=24-8\times1=16$ feet/second. $a(1)=-8$ feet/second².

Step6: Analyze when acceleration is non - zero and speed is increasing

Speed is increasing when $v(t)$ and $a(t)$ have the same sign. Since $a(t)=-8<0$ for all $t$, speed is increasing when $v(t)<0$, i.e., when $t>3$.

Step7: Graph the position function

The position function $s(t)=24t - 4t^{2}$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-4$, $b = 24$, $c = 0$. The vertex of the parabola $y = ax^{2}+bx + c$ has $t=-\frac{b}{2a}=-\frac{24}{2\times(-4)} = 3$, and $s(3)=24\times3-4\times3^{2}=72 - 36=36$. The $y$ - intercept is $s(0)=0$.

Answer:

a. The graph of the position function $s(t)=24t - 4t^{2}$ is a parabola opening downwards with vertex at $(3,36)$ and $y$ - intercept at $(0,0)$. The correct graph is the one that has a parabolic shape opening downwards. b. $v(t)=24 - 8t$. The object is stationary at $t = 3$ seconds, moving to the right when $0\leq t<3$ seconds, and moving to the left when $3<t\leq8$ seconds. c. $v(1)=16$ feet/second, $a(1)=-8$ feet/second². d. $a(t)=-8$ feet/second² for all $t$. e. The speed is increasing on the interval $(3,8]$.