Answer:

(a)

\(t=\frac{1}{gk}\ln\left(\frac{kU+\mu}{kv+\mu}\right).\)

(b)

\(25-\frac{25}{2}\ln3=11.3\text{ m}.\)

(c)

\(4.10\text{ m s}^{-1}.\)

(a) The resistance and friction both oppose the motion. Therefore Newton's second law gives

\(m\frac{dv}{dt}=-mgkv-\mu mg.\)

Cancel \(m\):

\(\frac{dv}{dt}=-g(kv+\mu).\)

Separate variables:

\(\frac{dv}{kv+\mu}=-g\,dt.\)

Integrating from \(v=U\) at \(t=0\) to speed \(v\) at time \(t\),

\(\int_U^v\frac{dV}{kV+\mu}=-g\int_0^t dt.\)

Thus

\(\frac1k\left[\ln(kV+\mu)\right]_U^v=-gt.\)

So

\(\frac1k\ln\left(\frac{kv+\mu}{kU+\mu}\right)=-gt.\)

Hence

\(t=\frac{1}{gk}\ln\left(\frac{kU+\mu}{kv+\mu}\right).\)

(b) Now \(U=10\), \(k=0.04\), \(\mu=0.2\), and \(g=10\). The equation of motion may be written as

\(v\frac{dv}{dx}=-10(0.04v+0.2).\)

So

\(v\frac{dv}{dx}=-\frac25v-2.\)

Separate variables:

\(dx=-\frac{v}{\frac25v+2}\,dv =-\frac52\frac{v}{v+5}\,dv.\)

Therefore

\(x=-\frac52\int\frac{v}{v+5}\,dv.\)

Since

\(\frac{v}{v+5}=1-\frac{5}{v+5},\)

we get

\(x=-\frac52\int\left(1-\frac5{v+5}\right)dv.\)

Thus

\(x=-\frac52v+\frac{25}{2}\ln(v+5)+C.\)

Initially \(x=0\) when \(v=10\), so

\(0=-25+\frac{25}{2}\ln15+C.\)

Hence

\(C=25-\frac{25}{2}\ln15.\)

The particle comes to rest when \(v=0\). Therefore

\(x=0+\frac{25}{2}\ln5+25-\frac{25}{2}\ln15.\)

So

\(x=25-\frac{25}{2}\ln3.\)

Numerically,

\(x\approx 11.3\text{ m}.\)

(c) From part (a), the time until the particle comes to rest is found by putting \(v=0\):

\(t=\frac{1}{10(0.04)}\ln\left(\frac{0.04(10)+0.2}{0.2}\right).\)

Thus

\(t=2.5\ln3.\)

The average speed is

\(\frac{\text{distance}}{\text{time}} =\frac{25-\frac{25}{2}\ln3}{2.5\ln3}.\)

Hence

\(\text{average speed}\approx 4.10\text{ m s}^{-1}.\)