A boat of mass $700 \mathrm{~kg}$ is travelling at a speed of $24 \mathrm{~ms}^{-1}$ when its engine is shut…
A boat of mass $700 \mathrm{~kg}$ is travelling at a speed of $24 \mathrm{~ms}^{-1}$ when its engine is shut off. The magnitude of frictional force $f$ between boat and water is given as $f=35 v$. Where $v$ is the speed in $\mathrm{ms}^{-1}$ and $f$ is in newton. The speed of boat becomes $6 \mathrm{~ms}^{-1}$ in time.
$18 \mathrm{~s}$
$36 \mathrm{~s}$
$34 \mathrm{~s}$
$28 \mathrm{~s}$
Solution
Given, Mass of boat, $m=700 \mathrm{~kg}$
Initial speed of boat, $v_1=24 \mathrm{~ms}^{-1}$
Final speed of boat, $v_2=6 \mathrm{~ms}^{-1}$
Frictional force, $f=35 \mathrm{v}$
According to Newton's second law,
$
f=-m \frac{d v}{d t}
$
$\begin{aligned} \Rightarrow \quad 35 v & =-m \frac{d v}{d t} \\ \Rightarrow \quad \frac{d v}{v} & =\frac{-35}{m} d t \\ \Rightarrow \quad \int_{v_1}^{v_1} \frac{d v}{v} & =\frac{-35}{m} \int_0^t d t \\ \ln \left(\frac{v_2}{v_1}\right) & =\frac{-35 t}{m} \\ \Rightarrow \quad t & =\frac{-m}{35} \ln \left(\frac{v_2}{v_1}\right)=\frac{-700}{35} \ln \left(\frac{6}{24}\right) \\ & =-20 \ln \frac{1}{4}=-20(\ln 1-\ln 4) \\ & =-20\left(0-\ln 2^2\right) \\ & =-20(-2 \ln 2) \\ & =40 \ln 2 \\ & =27.73 \\ & \simeq 28 \mathrm{~s}\end{aligned}$