Question

the position (in meters) of a marble, given an initial velocity and rolling up a long incline, is given by s = 120t / (1 + t), where t is measured in seconds and s = 0 is the starting point. a. graph the position function. choose the correct graph. a. graph with range 0,10 by -120,0 b. graph with range 0,10 by 0,120 c. graph with range 0,10 by -120,0 d. graph with range 0,10 by 0,120 b. find the velocity function for the marble. c. graph the velocity function and give a description of the motion of the marble. d. at what time is the marble 100 m from its starting point? e. at what time is the velocity 50 m/s?

Explanation:

Step1: Recall the properties of the position - velocity relationship

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=\frac{120t}{1 + t}$, we use the quotient - rule. The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 120t$, $u^\prime=120$, $v = 1 + t$, and $v^\prime = 1$.

Step2: Calculate the velocity function

$v(t)=\frac{120(1 + t)-120t\times1}{(1 + t)^{2}}=\frac{120+120t - 120t}{(1 + t)^{2}}=\frac{120}{(1 + t)^{2}}$.

Step3: Analyze the position function at $t = 0$

When $t = 0$, $s(0)=\frac{120\times0}{1+0}=0$. As $t$ increases, $s(t)=\frac{120t}{1 + t}=\frac{120(t + 1-1)}{1 + t}=120-\frac{120}{1 + t}$. The function $s(t)$ starts from $s = 0$ and increases towards $s = 120$ as $t
ightarrow\infty$.

Step4: Analyze the velocity function

The velocity function $v(t)=\frac{120}{(1 + t)^{2}}$ is always positive for $t\geq0$, and $v(t)$ is a decreasing function of $t$. When $t = 0$, $v(0)=120$ m/s, and as $t
ightarrow\infty$, $v(t)
ightarrow0$.

Step5: Solve for part (d)

Set $s(t)=100$. So, $\frac{120t}{1 + t}=100$. Cross - multiply: $120t=100(1 + t)$. Expand: $120t=100 + 100t$. Subtract $100t$ from both sides: $20t=100$, then $t = 5$ s.

Step6: Solve for part (e)

Set $v(t)=50$. So, $\frac{120}{(1 + t)^{2}}=50$. Cross - multiply: $120 = 50(1 + t)^{2}$. Then $(1 + t)^{2}=\frac{120}{50}=\frac{12}{5}$. Take the square root of both sides: $1 + t=\sqrt{\frac{12}{5}}\approx1.55$. So, $t=\sqrt{\frac{12}{5}}-1\approx0.55$ s.

a.

The position function $s(t)=\frac{120t}{1 + t}$. When $t = 0$, $s(0)=0$, and as $t
ightarrow\infty$, $s(t)
ightarrow120$. The correct graph is one that starts at the origin $(0,0)$ and increases towards $y = 120$. Looking at the options, assume the $x$ - axis is $t$ (from $0$ to some positive value) and the $y$ - axis is $s$. The graph should be increasing. Without seeing the exact details of the graphs, but based on the behavior of the function, if we consider the domain $t\in[0,10]$ and $s\in[0,120]$, the function starts at $(0,0)$ and increases.

b.

The velocity function $v(t)=\frac{120}{(1 + t)^{2}}$.

c.

The velocity function $v(t)=\frac{120}{(1 + t)^{2}}$ is a decreasing function of $t$ for $t\geq0$. The marble starts with an initial high velocity ($v(0) = 120$ m/s) and slows down as it moves up the incline, approaching a velocity of $0$ as $t
ightarrow\infty$.

d.

The time when the marble is 100 m from its starting point is $t = 5$ s.

e.

The time when the velocity is 50 m/s is $t=\sqrt{\frac{12}{5}}-1\approx0.55$ s.

Answer:

a. (No specific graph choice given as we don't have full details of the graphs, but it should start at $(0,0)$ and increase towards $y = 120$ in the domain $t\in[0,10]$ and $s\in[0,120]$)

b. $v(t)=\frac{120}{(1 + t)^{2}}$

c. The velocity is positive and decreasing, starting at $v(0)=120$ m/s and approaching $0$ as $t

ightarrow\infty$.

d. $t = 5$ s

e. $t=\sqrt{\frac{12}{5}}-1\approx0.55$ s