Λmotor boat of mass m moves in a lake with a velocity vo . Λt the moment t=0, the engine is shut

step 1 Hello, students in this question, we know that delta tb is equal to kb into m, right? So, y is e

Λmotor boat of mass m moves in a lake with a velocity vo ....

$\Lambda$ motor boat of mass $m$ moves in a lake with a velocity $v_{o} . \Lambda t$ the moment $t=0$, the engine is shut down. Assuming the resistance of water to be proportional to the velocity of the boat, $F=-r v(r$ is a positive constant), find : (a) how long the boat moved with the shut down engine, (b) the velocity of the boat as a function of the distance covered till the complete stop, and (c) the mean velocity of the boat over the time interval (beginning with the moment, $t=0$ ) during which the velocity decreases $\eta$ times. Solution Accelcration of the boat $--\frac{r v}{m}-\frac{d v}{d t}$ $\therefore \frac{d v}{v}--\frac{r}{m} d t$ Integrating, $\ln v-\frac{r}{m} t \mid C \quad$ When $t-0, v-v_{0}$ So, $C=\ln v_{0}$ $\therefore \quad \ln v--\frac{r}{m} t+\ln v_{o} \quad$ or $\quad \ln \frac{v}{v_{n}}--\frac{r}{m} t$ or $\frac{v}{v_{0}}=e^{\frac{r}{m} t}$ or $\quad v=v_{o} e^{\frac{r}{u x} t}$ i.e., $v \rightarrow 0$ when $t \rightarrow \infty$ So, it stops when $t$ is infinitc only. Also $\frac{v d v}{d x}--\frac{r v}{m} \quad$ or $\quad \frac{d v}{d x}--\frac{r}{m}$ $\therefore v--\frac{r}{m} x+D$ When $\quad x=0, v=v_{0} \quad \therefore \quad v_{\omega}=D \quad$ Henec $v--\frac{r}{m} x+v_{0} \quad \therefore \quad v-v_{\omega}-\frac{r x}{m}$ Total distanec it moves before it comes to stop is given by $v-v_{o} \quad \frac{r x}{m}-0 \quad \therefore \quad x-\frac{v_{o} m}{r}$ we have to find the distance $x^{\prime}$ covered before the velocity becomes $\frac{v_{0}}{\eta}$. Bocause $v=v_{o}-\frac{r x}{m} \quad \therefore \quad \frac{v_{0}}{\eta}=v_{o}-\frac{r x^{\prime}}{m} \quad$ or $\quad \frac{r x^{\prime}}{m}=v_{o}-\frac{v_{0}}{\eta}=v_{o}\left(1-\frac{1}{\eta}\right)$ $\therefore \quad x^{\prime}=\frac{m}{r}\left(\frac{\eta 1}{\eta}\right) v_{0}$ But $v-v_{0} e^{n} \quad \Rightarrow \frac{v_{0}}{v}-e^{n}$ $--e^{n t} \quad$ or $\quad \eta-e^{n}$ 246 (1)$\therefore \frac{r t}{m}-\ln e^{n} \quad \therefore \quad t-\frac{m \ln e^{\eta}}{r}$ $\Lambda$ verage velocity $-\frac{\text { distance travelled }}{\text { time }}-\frac{x^{\prime}}{t}-\frac{\frac{m}{r}\left(\frac{\eta}{\eta}\right) v_{o}}{m \ln e^{n}}-\frac{(\eta \quad \mathrm{l}) v_{a}}{\eta \ln e^{n}}$ 0

$\Lambda$ motor boat of mass $m$ moves in a lake with a velocity $v_{o} . \Lambda t$ the moment $t=0$, the engine is shut down. Assuming the resistance of water to be proportional to the velocity of the boat, $F=-r v(r$ is a positive constant), find :
(a) how long the boat moved with the shut down engine,
(b) the velocity of the boat as a function of the distance covered till the complete stop, and
(c) the mean velocity of the boat over the time interval (beginning with the moment, $t=0$ ) during which the velocity decreases $\eta$ times.
Solution
Accelcration of the boat $--\frac{r v}{m}-\frac{d v}{d t}$ $\therefore \frac{d v}{v}--\frac{r}{m} d t$
Integrating, $\ln v-\frac{r}{m} t \mid C \quad$ When $t-0, v-v_{0}$ So, $C=\ln v_{0}$
$\therefore \quad \ln v--\frac{r}{m} t+\ln v_{o} \quad$ or $\quad \ln \frac{v}{v_{n}}--\frac{r}{m} t$
or $\frac{v}{v_{0}}=e^{\frac{r}{m} t}$
or $\quad v=v_{o} e^{\frac{r}{u x} t}$ i.e., $v \rightarrow 0$ when $t \rightarrow \infty$
So, it stops when $t$ is infinitc only. Also $\frac{v d v}{d x}--\frac{r v}{m} \quad$ or $\quad \frac{d v}{d x}--\frac{r}{m}$
$\therefore v--\frac{r}{m} x+D$
When $\quad x=0, v=v_{0} \quad \therefore \quad v_{\omega}=D \quad$ Henec $v--\frac{r}{m} x+v_{0} \quad \therefore \quad v-v_{\omega}-\frac{r x}{m}$
Total distanec it moves before it comes to stop is given by
$v-v_{o} \quad \frac{r x}{m}-0 \quad \therefore \quad x-\frac{v_{o} m}{r}$
we have to find the distance $x^{\prime}$ covered before the velocity becomes $\frac{v_{0}}{\eta}$. Bocause $v=v_{o}-\frac{r x}{m} \quad \therefore \quad \frac{v_{0}}{\eta}=v_{o}-\frac{r x^{\prime}}{m} \quad$ or $\quad \frac{r x^{\prime}}{m}=v_{o}-\frac{v_{0}}{\eta}=v_{o}\left(1-\frac{1}{\eta}\right)$
$\therefore \quad x^{\prime}=\frac{m}{r}\left(\frac{\eta 1}{\eta}\right) v_{0}$
But $v-v_{0} e^{n} \quad \Rightarrow \frac{v_{0}}{v}-e^{n}$
$--e^{n t} \quad$ or $\quad \eta-e^{n}$
246
(1)$\therefore \frac{r t}{m}-\ln e^{n} \quad \therefore \quad t-\frac{m \ln e^{\eta}}{r}$
$\Lambda$ verage velocity $-\frac{\text { distance travelled }}{\text { time }}-\frac{x^{\prime}}{t}-\frac{\frac{m}{r}\left(\frac{\eta}{\eta}\right) v_{o}}{m \ln e^{n}}-\frac{(\eta \quad \mathrm{l}) v_{a}}{\eta \ln e^{n}}$
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Key Concepts

Average Value Calculation in Continuous Systems

Calculating the average (or mean) value of a varying quantity over a specific interval is an essential concept in analysis. This is typically done by integrating the quantity over the interval and dividing by the length of the interval. In problems involving deceleration, this method helps determine the overall performance, such as the mean velocity during a certain period, providing insights into the effective behavior of the system over time.

Chain Rule in Kinematics

The chain rule is a powerful tool in calculus used to relate the rates of change of variables that depend on other changing variables. In kinematics, it allows the conversion of a differential equation from being time-dependent to being position-dependent. By expressing acceleration as v(dv/dx), one can derive the relationship between velocity and distance, which is critical when the analysis requires relating distance traveled to the changing speed.

Separation of Variables in Differential Equations

Separation of variables is a method used to solve differential equations by isolating the different variables on separate sides of the equation. For instance, when the rate of change of velocity is proportional to the velocity itself, rewriting the equation in a form that allows integration with respect to each variable separately is fundamental to finding the solution, often leading to logarithmic relationships.

Newton's Second Law with Resistive Forces

This concept involves applying Newton's Second Law (F = ma) to systems where the force acting on an object includes a resistive term. In cases where the resistive force is proportional to velocity (F = -rv), the net force leads to a deceleration that can be modeled through a first-order differential equation, highlighting the interplay between the object's mass, its acceleration, and the damping force.

Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In dynamics, when an object experiences a resistive force proportional to its velocity, the solution to the resulting differential equation yields an exponential function. This function models how the velocity decreases over time, showing that the object never reaches exactly zero velocity in finite time but decays asymptotically.

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