Question

the position (in meters) of a marble, given an initial velocity and rolling up a long incline, is given by s = 160t / (t + 1), where t is measured in seconds and s = 0 is the starting point. a. graph the position function. choose the correct graph. 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 120 m from its starting point? e. at what time is the velocity 60 m/s?

Explanation:

Step1: Recall the relationship between position and velocity

Velocity $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=\frac{160t}{t + 1}$, 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, $u = 160t$, $u^\prime=160$, $v=t + 1$, and $v^\prime = 1$.
\[

$$\begin{align*} v(t)&=\frac{160(t + 1)-160t\times1}{(t + 1)^{2}}\\ &=\frac{160t+160 - 160t}{(t + 1)^{2}}\\ &=\frac{160}{(t + 1)^{2}} \end{align*}$$

\]

Step2: Analyze the position - function for graphing

When $t = 0$, $s(0)=\frac{160\times0}{0 + 1}=0$. As $t
ightarrow\infty$, $\lim_{t
ightarrow\infty}s(t)=\lim_{t
ightarrow\infty}\frac{160t}{t + 1}=\lim_{t
ightarrow\infty}\frac{160}{1+\frac{1}{t}} = 160$. The position - function $s(t)$ is a non - negative function for $t\geq0$ (since $t$ represents time), and it starts at $s(0) = 0$ and approaches $s = 160$ as $t$ gets large.

Step3: Analyze the velocity - function for graphing

The velocity function $v(t)=\frac{160}{(t + 1)^{2}}$ is also non - negative for $t\geq0$. When $t = 0$, $v(0)=\frac{160}{(0 + 1)^{2}}=160$. As $t
ightarrow\infty$, $\lim_{t
ightarrow\infty}v(t)=\lim_{t
ightarrow\infty}\frac{160}{(t + 1)^{2}}=0$.

Step4: Solve for when $s(t)=120$

Set $s(t)=120$, so $\frac{160t}{t + 1}=120$. Cross - multiply: $160t=120(t + 1)$. Expand: $160t=120t+120$. Subtract $120t$ from both sides: $160t-120t=120$, $40t=120$, and $t = 3$ seconds.

Step5: Solve for when $v(t)=60$

Set $v(t)=60$, so $\frac{160}{(t + 1)^{2}}=60$. Cross - multiply: $160 = 60(t + 1)^{2}$. Then $(t + 1)^{2}=\frac{160}{60}=\frac{8}{3}$. Take the square root of both sides: $t + 1=\sqrt{\frac{8}{3}}=\frac{2\sqrt{6}}{3}$ (we consider the positive root since $t\geq0$), and $t=\frac{2\sqrt{6}}{3}-1\approx\frac{2\times2.45}{3}-1=\frac{4.9}{3}-1\approx1.63 - 1=0.63$ seconds.

a. To graph $s(t)=\frac{160t}{t + 1}$, it starts at $(0,0)$ and approaches $y = 160$ as $t
ightarrow\infty$.
b. $v(t)=\frac{160}{(t + 1)^{2}}$, starts at $(0,160)$ and approaches $y = 0$ as $t
ightarrow\infty$.
c. The motion of the marble: It starts with an initial velocity of $160$ m/s and its velocity decreases over time as it moves up the incline, and its position increases from $0$ and approaches $160$ m.
d. Set $s(t)=120$, then $t = 3$ seconds.
e. Set $v(t)=60$, then $t=\frac{2\sqrt{6}}{3}-1\approx0.63$ seconds.

Answer:

a. The position - function $s(t)$ starts at $(0,0)$ and approaches $y = 160$ as $t
ightarrow\infty$.
b. $v(t)=\frac{160}{(t + 1)^{2}}$
c. The marble starts with an initial velocity of $160$ m/s, its velocity decreases, and its position increases from $0$ to approach $160$ m.
d. $t = 3$ seconds
e. $t=\frac{2\sqrt{6}}{3}-1\approx0.63$ seconds