Solution to Question 25
1. Understanding the Motion and Reference Frames
Let’s analyze the motion from the ground frame.
- The pedestrians are moving with the road at velocity $\vec{v}_p$. Its magnitude is $v$.
- The boy has an absolute velocity $\vec{v}_b$. We are given $|\vec{v}_b| = v$.
- To visit the pedestrians in sequence, the boy’s velocity relative to them, $\vec{v}_{rel} = \vec{v}_b – \vec{v}_p$, must follow the zigzag path shown in the diagram.
We need to find the distance the boy covers along the road. This is determined by the component of his absolute velocity $\vec{v}_b$ in the direction of the road, multiplied by the time $t$.
2. Geometry and Vector Relationships
Let $\theta$ be the angle the relative path makes with the direction of the road (horizontal). From the geometry of the pedestrians’ arrangement:
$$ \tan \theta = \frac{\text{vertical displacement}}{\text{horizontal displacement}} = \frac{b}{0.5a} = \frac{2b}{a} $$
The vector equation is $\vec{v}_b = \vec{v}_p + \vec{v}_{rel}$.
We are given that $|\vec{v}_b| = |\vec{v}_p| = v$. This means that in the vector diagram, $\vec{v}_b$ and $\vec{v}_p$ form two equal sides of an isosceles triangle, with $\vec{v}_{rel}$ as the third side connecting their heads.
3. Calculating the Net Velocity Along the Road
The boy must move “backwards” relative to the pedestrians to visit them in the correct sequence. This means $\vec{v}_{rel}$ makes an angle $\theta$ with the negative direction of the road.
In the isosceles triangle formed by $\vec{v}_b$ and $\vec{v}_p$, the angle bisector from their common origin is perpendicular to the vector difference $\vec{v}_{rel}$.
Let $\alpha$ be the angle between $\vec{v}_b$ and $\vec{v}_p$. The angle bisector is at $\alpha/2$.
The direction perpendicular to $\vec{v}_{rel}$ makes an angle of $90^\circ – \theta$ with the road’s direction. Therefore:
$$ \frac{\alpha}{2} = 90^\circ – \theta \implies \alpha = 180^\circ – 2\theta $$
The boy’s net speed along the road, $v_{net}$, is the component of $\vec{v}_b$ in the direction of $\vec{v}_p$:
$$ v_{net} = v \cos \alpha = v \cos(180^\circ – 2\theta) = -v \cos(2\theta) $$
Using the trigonometric identity $\cos(2\theta) = \frac{1 – \tan^2\theta}{1 + \tan^2\theta}$ and substituting $\tan\theta = \frac{2b}{a}$:
$$ v_{net} = -v \left( \frac{1 – (2b/a)^2}{1 + (2b/a)^2} \right) = -v \left( \frac{a^2 – 4b^2}{a^2 + 4b^2} \right) = v \left( \frac{4b^2 – a^2}{4b^2 + a^2} \right) $$
4. Numerical Calculation
Given values: $a = 2.0 \text{ m}$, $b = 3.0 \text{ m}$, $v = 1.5 \text{ m/s}$, $t = 2.0 \text{ min} = 120 \text{ s}$.
First, calculate the geometric factor:
$$ 4b^2 = 4(3.0)^2 = 36 $$
$$ a^2 = (2.0)^2 = 4 $$
$$ \frac{4b^2 – a^2}{4b^2 + a^2} = \frac{36 – 4}{36 + 4} = \frac{32}{40} = 0.8 $$
Now, calculate the net velocity and the total distance $l$:
$$ v_{net} = 1.5 \text{ m/s} \times 0.8 = 1.2 \text{ m/s} $$
$$ l = v_{net} \times t = 1.2 \text{ m/s} \times 120 \text{ s} = 144 \text{ m} $$
Final Answer: The distance covered by the boy along the road is 144 m.
Let’s analyze the motion from the ground frame.
- The pedestrians are moving with the road at velocity $\vec{v}_p$. Its magnitude is $v$.
- The boy has an absolute velocity $\vec{v}_b$. We are given $|\vec{v}_b| = v$.
- To visit the pedestrians in sequence, the boy’s velocity relative to them, $\vec{v}_{rel} = \vec{v}_b – \vec{v}_p$, must follow the zigzag path shown in the diagram.
We need to find the distance the boy covers along the road. This is determined by the component of his absolute velocity $\vec{v}_b$ in the direction of the road, multiplied by the time $t$.
Let $\theta$ be the angle the relative path makes with the direction of the road (horizontal). From the geometry of the pedestrians’ arrangement: $$ \tan \theta = \frac{\text{vertical displacement}}{\text{horizontal displacement}} = \frac{b}{0.5a} = \frac{2b}{a} $$ The vector equation is $\vec{v}_b = \vec{v}_p + \vec{v}_{rel}$.
We are given that $|\vec{v}_b| = |\vec{v}_p| = v$. This means that in the vector diagram, $\vec{v}_b$ and $\vec{v}_p$ form two equal sides of an isosceles triangle, with $\vec{v}_{rel}$ as the third side connecting their heads.
The boy must move “backwards” relative to the pedestrians to visit them in the correct sequence. This means $\vec{v}_{rel}$ makes an angle $\theta$ with the negative direction of the road.
In the isosceles triangle formed by $\vec{v}_b$ and $\vec{v}_p$, the angle bisector from their common origin is perpendicular to the vector difference $\vec{v}_{rel}$. Let $\alpha$ be the angle between $\vec{v}_b$ and $\vec{v}_p$. The angle bisector is at $\alpha/2$.
The direction perpendicular to $\vec{v}_{rel}$ makes an angle of $90^\circ – \theta$ with the road’s direction. Therefore: $$ \frac{\alpha}{2} = 90^\circ – \theta \implies \alpha = 180^\circ – 2\theta $$ The boy’s net speed along the road, $v_{net}$, is the component of $\vec{v}_b$ in the direction of $\vec{v}_p$: $$ v_{net} = v \cos \alpha = v \cos(180^\circ – 2\theta) = -v \cos(2\theta) $$ Using the trigonometric identity $\cos(2\theta) = \frac{1 – \tan^2\theta}{1 + \tan^2\theta}$ and substituting $\tan\theta = \frac{2b}{a}$: $$ v_{net} = -v \left( \frac{1 – (2b/a)^2}{1 + (2b/a)^2} \right) = -v \left( \frac{a^2 – 4b^2}{a^2 + 4b^2} \right) = v \left( \frac{4b^2 – a^2}{4b^2 + a^2} \right) $$
Given values: $a = 2.0 \text{ m}$, $b = 3.0 \text{ m}$, $v = 1.5 \text{ m/s}$, $t = 2.0 \text{ min} = 120 \text{ s}$.
First, calculate the geometric factor: $$ 4b^2 = 4(3.0)^2 = 36 $$ $$ a^2 = (2.0)^2 = 4 $$ $$ \frac{4b^2 – a^2}{4b^2 + a^2} = \frac{36 – 4}{36 + 4} = \frac{32}{40} = 0.8 $$
Now, calculate the net velocity and the total distance $l$: $$ v_{net} = 1.5 \text{ m/s} \times 0.8 = 1.2 \text{ m/s} $$ $$ l = v_{net} \times t = 1.2 \text{ m/s} \times 120 \text{ s} = 144 \text{ m} $$