Правка раздела «Answer»
en/1.2.3.md
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| ### Statement | |||
| $1.2.3.$ A body for time $t_0$ moves with constant velocity $v_0$. Then its velocity increases linearly with time so that at time $2 t_0$ it is equal to $2 v_0$. Determine the path traveled by the body for time $t > t_0$. | |||
|  | |||
| ### Solution | |||
| At time $t_0$, the coordinate was $x_1=v_0t_0$. | |||
| On the interval from $t_0$ to $2t_0$, the acceleration is constant and equal to $a=\frac{v_0}{t_0}$. | |||
| In the case of equiaxcelerated motion with initial velocity $v_0$ from time $t_0$, the path is found as | |||
| $$ | |||
| x_2 = v_0 (t-t_0)+\frac{a(t-t_0)^2}{2} | |||
| $$ | |||
| $$ | |||
| x = x_1+x_2 | |||
| $$ | |||
| $$ | |||
| x = v_0 t+\frac{a(t-t_0)^2}{2} | |||
| $$ | |||
| #### Answer | |||
| @@ -29,4 +29,3 @@Answer | |||
| $$ | |||
| L = v_{0}t + \frac{v_{0} (t − t_{0})^{2}}{2t_{0}} | |||
| $$ | |||
| − | Baha is gay 9c | ||
| ### Statement | ### Statement | ||
| $1.2.3.$ A body for time $t_0$ moves with constant velocity $v_0$. Then its velocity increases linearly with time so that at time $2 t_0$ it is equal to $2 v_0$. Determine the path traveled by the body for time $t > t_0$. | $1.2.3.$ A body for time $t_0$ moves with constant velocity $v_0$. Then its velocity increases linearly with time so that at time $2 t_0$ it is equal to $2 v_0$. Determine the path traveled by the body for time $t > t_0$. | ||
|  |  | ||
| ### Solution | ### Solution | ||
| At time $t_0$, the coordinate was $x_1=v_0t_0$. | At time $t_0$, the coordinate was $x_1=v_0t_0$. | ||
| On the interval from $t_0$ to $2t_0$, the acceleration is constant and equal to $a=\frac{v_0}{t_0}$. | On the interval from $t_0$ to $2t_0$, the acceleration is constant and equal to $a=\frac{v_0}{t_0}$. | ||
| In the case of equiaxcelerated motion with initial velocity $v_0$ from time $t_0$, the path is found as | In the case of equiaxcelerated motion with initial velocity $v_0$ from time $t_0$, the path is found as | ||
| $$ | $$ | ||
| x_2 = v_0 (t-t_0)+\frac{a(t-t_0)^2}{2} | x_2 = v_0 (t-t_0)+\frac{a(t-t_0)^2}{2} | ||
| $$ | $$ | ||
| $$ | $$ | ||
| x = x_1+x_2 | x = x_1+x_2 | ||
| $$ | $$ | ||
| $$ | $$ | ||
| x = v_0 t+\frac{a(t-t_0)^2}{2} | x = v_0 t+\frac{a(t-t_0)^2}{2} | ||
| $$ | $$ | ||
| #### Answer | #### Answer | ||
| @@ -29,4 +29,3 @@Answer | |||
| $$ | $$ | ||
| L = v_{0}t + \frac{v_{0} (t − t_{0})^{2}}{2t_{0}} | L = v_{0}t + \frac{v_{0} (t − t_{0})^{2}}{2t_{0}} | ||
| $$ | $$ | ||
| Baha is gay 9c | |||