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 360.5 meters before reaching a complete stop.
(round to the nearest tenth as needed.)
b. it will take \\(\square\\) seconds for the cyclists speed to drop to 4 m/sec.
(round to the nearest hundredth as needed.)

Explanation:

Step1: Find the mass \(m\)

The mass \(m\) of the cyclist and the bicycle is \(m = 24+14=38\) kg. The initial velocity \(v_0 = 37\) m/sec and \(k = 3.9\) kg/sec. The velocity function is \(v(t)=v_0e^{(-k/m)t}=37e^{(-3.9/38)t}\).

Step2: Solve for \(t\) when \(v = 4\)

We set \(v(t)=4\), so \(4 = 37e^{(-3.9/38)t}\).
First, divide both sides by 37: \(\frac{4}{37}=e^{(-3.9/38)t}\).
Then, take the natural logarithm of both sides: \(\ln(\frac{4}{37})=-\frac{3.9}{38}t\).
Now, solve for \(t\): \(t=-\frac{38}{3.9}\ln(\frac{4}{37})\).

Step3: Calculate the value of \(t\)

\(\ln(\frac{4}{37})\approx\ln(4)-\ln(37)\approx1.386 - 3.611=- 2.225\).
\(t =-\frac{38}{3.9}\times(- 2.225)=\frac{38\times2.225}{3.9}\).
\(38\times2.225 = 84.55\).
\(t=\frac{84.55}{3.9}\approx21.68\)

Answer:

It will take approximately \(21.68\) seconds.