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. 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? (round to the nearest integer as needed.) e. at what time is the velocity 60 m/s? the velocity is 60 m/s at t = s. give a description of the motion of the marble. choose the correct answer. a. the marble moves at a consistent speed until it stops at t = 10. b. the marble moves fastest at the beginning and slows considerably over the first 5 s. it continues to slow but never actually stops. c. the marble moves fastest at the beginning and slows considerably over the first 5 s. it continues to slow and stops at t = 10. d. the marble moves slowest at the beginning and moves fastest after the first 5 s. then it stops. the marble is 120 m from its starting point at t = 3 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 \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 160t\), \(u^\prime=160\), \(v=t + 1\), and \(v^\prime = 1\).
\[v(t)=\frac{160(t + 1)-160t\times1}{(t + 1)^{2}}=\frac{160t+160 - 160t}{(t + 1)^{2}}=\frac{160}{(t + 1)^{2}}\]

Step2: Solve for \(t\) when \(v(t)=60\)

Set \(\frac{160}{(t + 1)^{2}}=60\). Cross - multiply to get \(160 = 60(t + 1)^{2}\). Then \((t + 1)^{2}=\frac{160}{60}=\frac{8}{3}\). Take the square root of both sides: \(t + 1=\pm\sqrt{\frac{8}{3}}\). Since \(t\geq0\) (time cannot be negative in this context), \(t=\sqrt{\frac{8}{3}}-1\approx1.63 - 1=0.63\approx1\) s.

Step3: Solve for \(t\) when \(s(t)-s(0)=120\)

First, note that \(s(0) = 0\). 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: \(40t = 120\), and \(t = 3\) s.

Step4: Analyze the motion description

The velocity function \(v(t)=\frac{160}{(t + 1)^{2}}\) is a decreasing function for \(t\geq0\). The marble starts with \(v(0)=\frac{160}{(0 + 1)^{2}} = 160\) m/s and slows down over time. It moves fastest at the beginning and slows considerably over the first 5 s. It stops when \(v(t)=0\), and \(\lim_{t
ightarrow\infty}v(t)=0\). But for \(t\in[0,10]\), it stops when \(t = 10\) s. The correct description is that the marble moves fastest at the beginning and slows considerably over the first 5 s. It continues to slow and stops at \(t = 10\) s.

Answer:

a. The position function \(s(t)=\frac{160t}{t + 1}\) can be graphed using a graphing utility. The general shape is an increasing function that starts at the origin \((0,0)\) and approaches a horizontal asymptote \(y = 160\) as \(t
ightarrow\infty\).
b. The velocity function \(v(t)=\frac{160}{(t + 1)^{2}}\). It can be graphed as a decreasing function starting at \(v(0)=160\) m/s and approaching \(v = 0\) as \(t
ightarrow\infty\).
c. The marble moves fastest at the beginning and slows considerably over the first 5 s. It continues to slow and stops at \(t = 10\) s.
d. At \(t = 3\) s, the marble is 120 m from its starting point.
e. At \(t\approx1\) s, the velocity is 60 m/s.