Question

a 24-kg cyclist on a 14-kg bicycle starts coasting on level ground at 37 m/sec. use $v = v_0 e^{(-k/m)t}$, where k is about 3.9 kg/sec.
a. about how far will the cyclist coast before reaching a complete stop?
b. how long will it take the cyclists speed to drop to 4 m/sec?
a. the cyclist will coast \\(\square\\) meters before reaching a complete stop.
(round to the nearest tenth as needed.)

Explanation:

Step 1: Calculate the total mass \(m\)

The total mass \(m\) of the cyclist and the bicycle is \(m = 24+14=38\) kg. The initial velocity \(v_0 = 37\) m/s and \(k = 3.9\) kg/s. The velocity function is \(v(t)=v_0e^{-(k/m)t}=37e^{-(3.9/38)t}\). The distance \(s(t)\) is given by the integral of \(v(t)\) from \(t = 0\) to \(t=\infty\). Using the formula \(\int_{0}^{\infty}v_0e^{-(k/m)t}dt\).
We know that \(\int_{0}^{\infty}ae^{-bt}dt=\frac{a}{b}\) (where \(a = v_0\) and \(b=\frac{k}{m}\)).
Substitute \(a = 37\), \(b=\frac{3.9}{38}\) into the formula.

Step 2: Compute the distance

$$
LATEXBLOCK0
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Answer:

\(360.5\)