Правка разделов «Statement», «Solution», «Answer»
en/7.4.18.md
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| @@ -1,44 +1,17 @@ | |||
| ### Statement | |||
| − | $7.4.18.$ [Insert problem description here] | ||
| + | $7.4.18.$ Two identical particles with charge $q$ and velocities $u$ and $v$ lying in the same plane, making up angles $\alpha$ and $\beta$ with a line connecting them, respectively, are located at a distance $l$ from each other. Determine the mass of the particles if it is known that the minimum distance to which they approach is equal to $r$. | ||
| − | __Example Statement__: | ||
| − | $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. | ||
| − | |||
| − | |||
| ### Solution | |||
| − | [Your solution should be placed here] | ||
| + |  | ||
| − | __Example Solution__: | ||
| − | The acceleration of the body defined by | ||
| + |  | ||
| − | $$a(t) = bt$$ | ||
| + |  | ||
| − | We know that acceleration is the time derivative of velocity: | ||
| − | |||
| − | $$a(t) = \frac{d v(t)}{d t}$$ | ||
| − | |||
| − | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | ||
| − | |||
| − | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | ||
| − | |||
| − | If the initial velocity is $v(0) = 0$, then the velocity becomes: | ||
| − | |||
| − | $$v(t) = \frac{b t^2}{2}$$ | ||
| − | |||
| − | Likewise, integrate $v(t)$ with respect to time: | ||
| − | |||
| − | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | ||
| − | |||
| − | From where the coordinate from time, considering the initial conditions: | ||
| − | |||
| − | $$\boxed{x(t)=\frac{bt^3}{6}}$$ | ||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result, like this:] | ||
| − | |||
| − | |||
| − | __Example Answer__: | ||
| − | $$ x(t)=\frac{bt^3}{6} $$ | ||
| + | $$ | ||
| + | m =\frac{4q^2(l-r)}{rl\left[u^2+v^2+2uv\cos(\alpha+\beta)-\frac{l^2}{r^2}(u\sin\alpha-v\sin\beta)^2\right]} | ||
| + | $$ | ||
| @@ -1,44 +1,17 @@ | |||
| ### Statement | ### Statement | ||
| $7.4.18.$ [Insert problem description here] | $7.4.18.$ Two identical particles with charge $q$ and velocities $u$ and $v$ lying in the same plane, making up angles $\alpha$ and $\beta$ with a line connecting them, respectively, are located at a distance $l$ from each other. Determine the mass of the particles if it is known that the minimum distance to which they approach is equal to $r$. | ||
| __Example Statement__: | |||
| $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. | |||
| ### Solution | ### Solution | ||
| [Your solution should be placed here] |  | ||
| __Example Solution__: |  | ||
| The acceleration of the body defined by | |||
| $$a(t) = bt$$ |  | ||
| We know that acceleration is the time derivative of velocity: | |||
| $$a(t) = \frac{d v(t)}{d t}$$ | |||
| To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | |||
| $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | |||
| If the initial velocity is $v(0) = 0$, then the velocity becomes: | |||
| $$v(t) = \frac{b t^2}{2}$$ | |||
| Likewise, integrate $v(t)$ with respect to time: | |||
| $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | |||
| From where the coordinate from time, considering the initial conditions: | |||
| $$\boxed{x(t)=\frac{bt^3}{6}}$$ | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result, like this:] | $$ | ||
| m =\frac{4q^2(l-r)}{rl\left[u^2+v^2+2uv\cos(\alpha+\beta)-\frac{l^2}{r^2}(u\sin\alpha-v\sin\beta)^2\right]} | |||
| $$ | |||
| __Example Answer__: | |||
| $$ x(t)=\frac{bt^3}{6} $$ | |||