Правка разделов «Solution», «Answer»
en/1.2.16.md
+4 −4
| ### Statement | |||
| $1.2.16.$ | |||
| The graphs of the coordinate dependence on time is plotted for two particles using two different time | |||
| scales. The plotted graphs turn out to be the same. One division of the time axis | |||
| $t$ for the graph of the first particle corresponds to $4 \text{s}$, and for the graph of the | |||
| second particle is $1 \text{s}$. Find the ratio of velocities and the ratio of particle accelerations | |||
| for point A of the graph. | |||
| ### Solution | |||
| We know the instantaneous velocity and acceleration of any object at a given point can be extracted from a plot using | |||
| \begin{align*} | |||
| v &= \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{d x}{d t} | |||
| \end{align*} | |||
| @@ -15,7 +15,7 @@Solution | |||
| \begin{align*} | |||
| a &= \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} =\frac{d v }{d t} | |||
| \end{align*} | |||
| − | We note that, for any given time, the graphs exactly coincide, so any $\Delta x$ is exactly the same for both particles, and so $dx_1 = dx_2$. Likewise, for any time interval, we know that $\Delta t_1 = 4 \Delta t_2$, so $dt_1 = 4dt_2$. We can then say for | ||
| + | We note that, for any given time, the graphs exactly coincide, so any $\Delta x$ is exactly the same for both particles, and so $dx_1 = dx_2$. Likewise, for any time interval, we know that $\Delta t_1 = 4 \Delta t_2$, so $dt_1 = 4dt_2$. We can then say, for an equivalent Point A on their graphs, that: | ||
| \begin{align*} | |||
| v_1 &=\frac{d x_1}{d t_1} = \frac{d x_2}{4 d t_2} = \frac{1}{4} \left( \frac{d x_2}{d t_2} \right) = \frac{1}{4} v_2 | |||
| \end{align*} | |||
| More generally, we can say that, if the time scale divisions are related by $\Delta t_1 / \Delta t_2 = \alpha$, the ratio of velocities are given by: | |||
| \begin{align*} | |||
| \frac{v_2}{v_1} = \alpha | |||
| \end{align*} | |||
| Using the same logic, we can then say for the accelerations that: | |||
| @@ -27,23 +27,23 @@Solution | |||
| \begin{align*} | |||
| a_1 &= \frac{d v_1 }{d t_1} = \frac{d \left( \frac{1}{4} v_2 \right)}{4 d t_2} = \frac{1}{16} \left( \frac{d v_2}{d t_2} \right) = \frac{1}{16} a_2 | |||
| \end{align*} | |||
| − | More generally, the ratio of | ||
| + | More generally, the ratio of accelerations is given by: | ||
| \begin{align*} | |||
| \frac{a_2}{a_1} = \alpha^2 | |||
| \end{align*} | |||
| #### Answer | |||
| − | The ratio of velocities is | ||
| + | The ratio of velocities is | ||
| \begin{align} | |||
| \boxed{ | |||
| \frac{v_2}{v_1} = 4 | |||
| } | |||
| \end{align} | |||
| − | And the ratio of the acceleration is | ||
| + | And the ratio of the acceleration is | ||
| \begin{align} | |||
| \boxed{ | |||
| \frac{a_2}{a_1} = 4^2 = 16 | |||
| } | |||
| \end{align} | |||
| ### Statement | ### Statement | ||
| $1.2.16.$ | $1.2.16.$ | ||
| The graphs of the coordinate dependence on time is plotted for two particles using two different time | The graphs of the coordinate dependence on time is plotted for two particles using two different time | ||
| scales. The plotted graphs turn out to be the same. One division of the time axis | scales. The plotted graphs turn out to be the same. One division of the time axis | ||
| $t$ for the graph of the first particle corresponds to $4 \text{s}$, and for the graph of the | $t$ for the graph of the first particle corresponds to $4 \text{s}$, and for the graph of the | ||
| second particle is $1 \text{s}$. Find the ratio of velocities and the ratio of particle accelerations | second particle is $1 \text{s}$. Find the ratio of velocities and the ratio of particle accelerations | ||
| for point A of the graph. | for point A of the graph. | ||
| ### Solution | ### Solution | ||
| We know the instantaneous velocity and acceleration of any object at a given point can be extracted from a plot using | We know the instantaneous velocity and acceleration of any object at a given point can be extracted from a plot using | ||
| \begin{align*} | \begin{align*} | ||
| v &= \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{d x}{d t} | v &= \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{d x}{d t} | ||
| \end{align*} | \end{align*} | ||
| @@ -15,7 +15,7 @@Solution | |||
| \begin{align*} | \begin{align*} | ||
| a &= \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} =\frac{d v }{d t} | a &= \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} =\frac{d v }{d t} | ||
| \end{align*} | \end{align*} | ||
| We note that, for any given time, the graphs exactly coincide, so any $\Delta x$ is exactly the same for both particles, and so $dx_1 = dx_2$. Likewise, for any time interval, we know that $\Delta t_1 = 4 \Delta t_2$, so $dt_1 = 4dt_2$. We can then say for |
We note that, for any given time, the graphs exactly coincide, so any $\Delta x$ is exactly the same for both particles, and so $dx_1 = dx_2$. Likewise, for any time interval, we know that $\Delta t_1 = 4 \Delta t_2$, so $dt_1 = 4dt_2$. We can then say, for an equivalent Point A on their graphs, that: | ||
| \begin{align*} | \begin{align*} | ||
| v_1 &=\frac{d x_1}{d t_1} = \frac{d x_2}{4 d t_2} = \frac{1}{4} \left( \frac{d x_2}{d t_2} \right) = \frac{1}{4} v_2 | v_1 &=\frac{d x_1}{d t_1} = \frac{d x_2}{4 d t_2} = \frac{1}{4} \left( \frac{d x_2}{d t_2} \right) = \frac{1}{4} v_2 | ||
| \end{align*} | \end{align*} | ||
| More generally, we can say that, if the time scale divisions are related by $\Delta t_1 / \Delta t_2 = \alpha$, the ratio of velocities are given by: | More generally, we can say that, if the time scale divisions are related by $\Delta t_1 / \Delta t_2 = \alpha$, the ratio of velocities are given by: | ||
| \begin{align*} | \begin{align*} | ||
| \frac{v_2}{v_1} = \alpha | \frac{v_2}{v_1} = \alpha | ||
| \end{align*} | \end{align*} | ||
| Using the same logic, we can then say for the accelerations that: | Using the same logic, we can then say for the accelerations that: | ||
| @@ -27,23 +27,23 @@Solution | |||
| \begin{align*} | \begin{align*} | ||
| a_1 &= \frac{d v_1 }{d t_1} = \frac{d \left( \frac{1}{4} v_2 \right)}{4 d t_2} = \frac{1}{16} \left( \frac{d v_2}{d t_2} \right) = \frac{1}{16} a_2 | a_1 &= \frac{d v_1 }{d t_1} = \frac{d \left( \frac{1}{4} v_2 \right)}{4 d t_2} = \frac{1}{16} \left( \frac{d v_2}{d t_2} \right) = \frac{1}{16} a_2 | ||
| \end{align*} | \end{align*} | ||
| More generally, the ratio of |
More generally, the ratio of accelerations is given by: | ||
| \begin{align*} | \begin{align*} | ||
| \frac{a_2}{a_1} = \alpha^2 | \frac{a_2}{a_1} = \alpha^2 | ||
| \end{align*} | \end{align*} | ||
| #### Answer | #### Answer | ||
| The ratio of velocities is |
The ratio of velocities is | ||
| \begin{align} | \begin{align} | ||
| \boxed{ | \boxed{ | ||
| \frac{v_2}{v_1} = 4 | \frac{v_2}{v_1} = 4 | ||
| } | } | ||
| \end{align} | \end{align} | ||
| And the ratio of the acceleration is |
And the ratio of the acceleration is | ||
| \begin{align} | \begin{align} | ||
| \boxed{ | \boxed{ | ||
| \frac{a_2}{a_1} = 4^2 = 16 | \frac{a_2}{a_1} = 4^2 = 16 | ||
| } | } | ||
| \end{align} | \end{align} | ||