Question

the position (in meters) of a marble, given an initial velocity and rolling up a long incline, is given by s = 120t / (t + 1), where t is measured in seconds and s = 0 is the starting point. a. graph the position function. choose the correct graph. a. 0,10 by -120,0 b. 0,10 by 0,120 c. 0,10 by -120,0 d. 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:

1. Find the velocity function:

• The position - function is given by \(s(t)=\frac{120t}{t + 1}\), where \(t\geq0\).
• The velocity function \(v(t)\) is the derivative of the position function. Using the quotient - rule, if \(s(t)=\frac{u(t)}{v(t)}\) where \(u(t)=120t\) and \(v(t)=t + 1\), the quotient - rule states that \(s^\prime(t)=\frac{u^\prime(t)v(t)-u(t)v^\prime(t)}{v(t)^2}\).
• First, find \(u^\prime(t)\) and \(v^\prime(t)\): \(u^\prime(t)=120\) and \(v^\prime(t)=1\).
• Then, \(v(t)=\frac{120(t + 1)-120t}{(t + 1)^2}=\frac{120t+120 - 120t}{(t + 1)^2}=\frac{120}{(t + 1)^2}\).

1. a. Graph the position function:

• When \(t = 0\), \(s(0)=\frac{120\times0}{0 + 1}=0\).
• As \(t\to\infty\), \(\lim_{t\to\infty}\frac{120t}{t + 1}=\lim_{t\to\infty}\frac{120}{1+\frac{1}{t}} = 120\). The graph of \(y = s(t)\) starts at the origin \((0,0)\) and approaches \(y = 120\) as \(t\) increases.

1. b. Find the velocity function:

• We already found that \(v(t)=\frac{120}{(t + 1)^2}\).

1. c. Graph the velocity function:

• When \(t = 0\), \(v(0)=\frac{120}{(0 + 1)^2}=120\).
• As \(t\to\infty\), \(\lim_{t\to\infty}\frac{120}{(t + 1)^2}=0\). The graph of \(y = v(t)\) starts at \((0,120)\) and approaches \(y = 0\) as \(t\) increases. The motion of the marble is that it starts with an initial velocity of \(120\ m/s\) and its velocity decreases over time as it moves up the incline.

1. d. At what time is the marble \(100\ m\) from its starting point?

• Set \(s(t)=100\), so \(\frac{120t}{t + 1}=100\).
• Cross - multiply: \(120t=100(t + 1)\).
• Expand: \(120t=100t + 100\).
• Subtract \(100t\) from both sides: \(120t-100t=100\), \(20t=100\).
• Solve for \(t\): \(t = 5\) seconds.

1. e. At what time is the velocity \(50\ m/s\)?

• Set \(v(t)=50\), so \(\frac{120}{(t + 1)^2}=50\).
• Cross - multiply: \(120 = 50(t + 1)^2\).
• Divide both sides by \(50\): \(\frac{120}{50}=(t + 1)^2\), \(\frac{12}{5}=(t + 1)^2\).
• Take the square root of both sides: \(t + 1=\pm\sqrt{\frac{12}{5}}\). Since \(t\geq0\), \(t + 1=\sqrt{\frac{12}{5}}\approx\sqrt{2.4}\approx1.55\).
• Then \(t=\sqrt{\frac{12}{5}}-1\approx1.55 - 1 = 0.55\) seconds.

Step1: Find the velocity function

Using the quotient - rule on \(s(t)=\frac{120t}{t + 1}\), we get \(v(t)=\frac{120}{(t + 1)^2}\).

Step2: Graph the position function

Analyze \(s(0) = 0\) and \(\lim_{t\to\infty}s(t)=120\) to graph \(s(t)\).

Step3: Graph the velocity function

Analyze \(v(0)=120\) and \(\lim_{t\to\infty}v(t)=0\) to graph \(v(t)\).

Step4: Solve for \(t\) when \(s(t)=100\)

Set \(\frac{120t}{t + 1}=100\) and solve for \(t\) to get \(t = 5\) seconds.

Step5: Solve for \(t\) when \(v(t)=50\)

Set \(\frac{120}{(t + 1)^2}=50\) and solve for \(t\) to get \(t=\sqrt{\frac{12}{5}}-1\approx0.55\) seconds.

Answer:

a. The graph of \(s(t)=\frac{120t}{t + 1}\) starts at \((0,0)\) and approaches \(y = 120\) as \(t\) increases.
b. \(v(t)=\frac{120}{(t + 1)^2}\).
c. The graph of \(v(t)\) starts at \((0,120)\) and approaches \(y = 0\) as \(t\) increases. The marble starts with an initial velocity of \(120\ m/s\) and its velocity decreases over time.
d. \(t = 5\) seconds.
e. \(t=\sqrt{\frac{12}{5}}-1\approx0.55\) seconds.