Edit to “Solution”
en/1.2.5.md
+2 −2
| ### Statement | |||
| $1.2.5.$ When entering a damaged section of the highway, each car in a convoy reduces its speed from $v_1$ to $v_2$. What should be the distance between the cars so that they do not collide? The length of each car is $l$. | |||
| ### Solution | |||
| Let's move to the frame of reference associated with the $1$th car at the moment when it enters the damaged area and starts moving with velocity $v_2$. The second car will move with a velocity equal to the velocity difference $$v_\text{from}=v_1-v_2$$ | |||
| @@ -8,13 +8,13 @@Solution | |||
| The time taken by the second car in the reference frame of the "stopped" first car is given by: | |||
| − | $$t=\frac{L}{v_1–v_2}, | ||
| + | $$t=\frac{L}{v_1–v_2},\tag{1}$$ | ||
| where $L$ is the distance between the cars. | |||
| During the time $t$, the first car, moving with the modified speed $v_2$, must cover at least a distance equal to its own length: | |||
| − | $$t=\frac{l}{v_2} \ | ||
| + | $$t=\frac{l}{v_2} \tag{2}$$ | ||
| Equating $(1)$ and $(2)$: | |||
| $$\frac{L}{v_1–v_2}=\frac{l}{v_2}$$ | |||
| From this, the minimum distance $L$ between the cars, such that they do not collide, is given by: | |||
| $$\fbox{$L \geq \frac{v_1–v_2}{v_2} \cdot l$}$$ | |||
| #### Answer | |||
| $$L=\frac{v_1–v_2}{v_2} \cdot l$$ | |||
| unchanged lines 9 | |||
| ### Statement | ### Statement | ||
| $1.2.5.$ When entering a damaged section of the highway, each car in a convoy reduces its speed from $v_1$ to $v_2$. What should be the distance between the cars so that they do not collide? The length of each car is $l$. | $1.2.5.$ When entering a damaged section of the highway, each car in a convoy reduces its speed from $v_1$ to $v_2$. What should be the distance between the cars so that they do not collide? The length of each car is $l$. | ||
| ### Solution | ### Solution | ||
| Let's move to the frame of reference associated with the $1$th car at the moment when it enters the damaged area and starts moving with velocity $v_2$. The second car will move with a velocity equal to the velocity difference $$v_\text{from}=v_1-v_2$$ | Let's move to the frame of reference associated with the $1$th car at the moment when it enters the damaged area and starts moving with velocity $v_2$. The second car will move with a velocity equal to the velocity difference $$v_\text{from}=v_1-v_2$$ | ||
| @@ -8,13 +8,13 @@Solution | |||
| The time taken by the second car in the reference frame of the "stopped" first car is given by: | The time taken by the second car in the reference frame of the "stopped" first car is given by: | ||
| $$t=\frac{L}{v_1–v_2}, |
$$t=\frac{L}{v_1–v_2},\tag{1}$$ | ||
| where $L$ is the distance between the cars. | where $L$ is the distance between the cars. | ||
| During the time $t$, the first car, moving with the modified speed $v_2$, must cover at least a distance equal to its own length: | During the time $t$, the first car, moving with the modified speed $v_2$, must cover at least a distance equal to its own length: | ||
| $$t=\frac{l}{v_2} \ |
$$t=\frac{l}{v_2} \tag{2}$$ | ||
| Equating $(1)$ and $(2)$: | Equating $(1)$ and $(2)$: | ||
| $$\frac{L}{v_1–v_2}=\frac{l}{v_2}$$ | $$\frac{L}{v_1–v_2}=\frac{l}{v_2}$$ | ||
| From this, the minimum distance $L$ between the cars, such that they do not collide, is given by: | From this, the minimum distance $L$ between the cars, such that they do not collide, is given by: | ||
| $$\fbox{$L \geq \frac{v_1–v_2}{v_2} \cdot l$}$$ | $$\fbox{$L \geq \frac{v_1–v_2}{v_2} \cdot l$}$$ | ||
| #### Answer | #### Answer | ||
| $$L=\frac{v_1–v_2}{v_2} \cdot l$$ | $$L=\frac{v_1–v_2}{v_2} \cdot l$$ | ||
| unchanged lines 9 | |||