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| ### Statement |
| ### Statement |
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| $2.4.42.$ [Insert the problem statement] |
| $2.4.42.$ The power of a car with an electric motor depends on the angular velocity of rotation of the wheels according to the law $(A-B\omega)\omega$. The vehicle's steady speed on a horizontal highway is $70$ km/hr. Without a load, it can overcome ascents with an angle of inclination of the highway up to $45^\circ$. What is the steady-state speed of the car when climbing with a highway slope of $30^\circ$? What kind of climbs can it overcome when the weight of the load is equal to the mass of the car? |
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| ### Solution |
| ### Solution |
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| If the wheels roll without slipping, then $\omega$ varies directly with the speed $v$ of the car (in km/hr). Thus, the power of the motor can be expressed as $P(v)=(\alpha-\beta v)v$. (We can use an appropriate unit of force for the terms in parentheses and consequently an appropriate unit of power for $P(v)$.) If air resistance is neglected, then the speed of the car being $70$ km/hr on a horizontal highway means that $P(70)=0$, and so $\alpha=70\beta$. Let |
| If the wheels of the car roll without slipping, then $\omega$ varies directly with the speed $v$ of the car (in km/hr). Thus, the power of the motor can be expressed as $P(v)=(\alpha-\beta v)v$. (We can use an appropriate unit of force for the terms in parentheses and consequently an appropriate unit of power for $P$.) If air resistance is neglected, then the speed of the car being $70$ km/hr on a horizontal highway means that $P(70)=0$, and so $\alpha=70\beta$. Let |
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| \[F(v)=P(v)/v=\beta(70-v)\] |
| \[F(v)=P(v)/v=\beta(70-v)\] |
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| denote the driving force that the motor gives to the car. The fact that the car can barely climb an inclination of $45^\circ$ means that $F(0)=W\sin45^\circ$, where $W$ is the weight of the car only and $W\sin45^\circ$ is the component of the weight of the car down the incline. It follows that $\beta=W/(70\sqrt2)$. To find the speed of the car going up an inclination of $30^\circ$, we solve for $v$ from the equation $F(v)=W\sin30^\circ$; that is, |
| denote the equivalent driving force that the motor gives to the car. The fact that the car can barely climb an inclination of $45^\circ$ means that $F(0)=W\sin45^\circ$, where $W$ is the weight of the car, and $W\sin45^\circ$ is the component of the weight of the car down the incline. It follows that $\beta=W/(70\sqrt2)$. To find the speed of the car going up an inclination of $30^\circ$, we solve for $v$ from the equation $F(v)=W\sin30^\circ$; that is, |
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| \[\frac{W}{70\sqrt2}(70-v)=\frac{W}{2}\qquad\Rightarrow\qquad v=70-35\sqrt2,\] |
| \[\frac{W}{70\sqrt2}\cdot(70-v)=\frac{W}{2}\qquad\Rightarrow\qquad v=70-35\sqrt2,\] |
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| which is approximately $20.5$ km/hr. To find the maximum inclination that the car with a load equal to its weight can climb, we solve for $\theta$ from the equation $F(0)=2W\sin\theta$; that is, |
| which is approximately $20.5$ km/hr. To find the maximum inclination that the car with a load equal to its weight can climb, we solve for $\theta$ from the equation $F(0)=2W\sin\theta$; that is, |
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| \[2W\sin\theta=W\sin45^\circ\qquad\Rightarrow\qquad\theta=\arcsin\frac{1}{2\sqrt2}.\] |
| \[2W\sin\theta=W\sin45^\circ\qquad\Rightarrow\qquad\theta=\arcsin\frac{\sqrt2}{4}.\] |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $v=70-35\sqrt2\approx20.5$ km/hr |
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| $\theta=\arcsin\frac{\sqrt2}{4}$ |
