Edit to “Solution”
en/14.4.4.md
+2 −2
| ### Statement | |||
| $14.4.4.$ [Insert problem description here] | |||
| __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. | |||
| @@ -8,8 +8,8 @@Statement | |||
| ### Solution | |||
| − | [Your solution should be placed here] | ||
| − | d | ||
| + | |||
| + | |||
| __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__: | |||
| $$ x(t)=\frac{bt^3}{6} $$ | |||
| unchanged lines 29 | |||
| ### Statement | ### Statement | ||
| $14.4.4.$ [Insert problem description here] | $14.4.4.$ [Insert problem description here] | ||
| __Example Statement__: | __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. | $1.1.1.$ 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. | ||
| @@ -8,8 +8,8 @@Statement | |||
| ### Solution | ### Solution | ||
| [Your solution should be placed here] | |||
| d | |||
| __Example Solution__: | __Example Solution__: | ||
| The acceleration of the body defined by | The acceleration of the body defined by | ||
| $$a(t) = bt$$ | $$a(t) = bt$$ | ||
| We know that acceleration is the time derivative of velocity: | We know that acceleration is the time derivative of velocity: | ||
| $$a(t) = \frac{d v(t)}{d t}$$ | $$a(t) = \frac{d v(t)}{d t}$$ | ||
| To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | ||
| $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | ||
| If the initial velocity is $v(0) = 0$, then the velocity becomes: | If the initial velocity is $v(0) = 0$, then the velocity becomes: | ||
| $$v(t) = \frac{b t^2}{2}$$ | $$v(t) = \frac{b t^2}{2}$$ | ||
| Likewise, integrate $v(t)$ with respect to time: | Likewise, integrate $v(t)$ with respect to time: | ||
| $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | ||
| From where the coordinate from time, considering the initial conditions: | From where the coordinate from time, considering the initial conditions: | ||
| $$\boxed{x(t)=\frac{bt^3}{6}}$$ | $$\boxed{x(t)=\frac{bt^3}{6}}$$ | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result, like this:] | [Insert a concise answer or boxed result, like this:] | ||
| __Example Answer__: | __Example Answer__: | ||
| $$ x(t)=\frac{bt^3}{6} $$ | $$ x(t)=\frac{bt^3}{6} $$ | ||
| unchanged lines 29 | |||