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en/2.4.42.md
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| + | ### Statement | ||
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| + | $2.4.42.$ [Insert the problem statement] | ||
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| + | ### 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 | ||
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| + | \[F(v)=P(v)/v=\beta(70-v)\] | ||
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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, | ||
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| + | \[\frac{W}{70\sqrt2}(70-v)=\frac{W}{2}\qquad\Rightarrow\qquad v=70-35\sqrt2,\] | ||
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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, | ||
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| + | \[2W\sin\theta=W\sin45^\circ\qquad\Rightarrow\qquad\theta=\arcsin\frac{1}{2\sqrt2}.\] | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.4.42.$ [Insert the problem statement] | |||
| ### Solution | |||
| 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 | |||
| \[F(v)=P(v)/v=\beta(70-v)\] | |||
| 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, | |||
| \[\frac{W}{70\sqrt2}(70-v)=\frac{W}{2}\qquad\Rightarrow\qquad v=70-35\sqrt2,\] | |||
| 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, | |||
| \[2W\sin\theta=W\sin45^\circ\qquad\Rightarrow\qquad\theta=\arcsin\frac{1}{2\sqrt2}.\] | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||