On a large slippery ground, a boy left his dog sitting and walks away with a constant velocity v_b =

Q: Why does the dog need to decelerate? Can't it just keep accelerating until it catches the boy?

The problem states the dog must "catch him and thereafter move together." If the dog kept accelerating, it would be moving much faster than the boy when it reaches him, meaning it would overshoot or collide rather than moving together smoothly. It must match the boy's velocity exactly at the moment of catching up.

Q: Why is the area between the two curves equal to the initial separation $x_0$?

The area under a v-t graph is the displacement. The area under the dog's curve is its total displacement $x_d$, and the area under the boy's curve is his displacement $v_b t$. Since the dog starts $x_0$ distance behind, $x_d = x_0 + v_b t$. Rearranging gives $x_d - v_b t = x_0$, which is exactly the area between the two curves.

Q: How do we know the acceleration and deceleration times are related?

The dog accelerates at $+a$ and decelerates at $-a$. Because the magnitudes of the slopes are equal, the time taken to change velocity by a certain amount is the same whether accelerating or decelerating. This symmetry makes the green triangle on our v-t graph an isosceles triangle if we scale the axes equally.

Q: What does the orange area $A_2$ represent physically?

$A_2$ represents the extra distance the boy travels away from the dog during the initial phase when the dog's speed is still less than the boy's speed. The dog is actually losing ground during this time!

Q: Can we solve this using standard kinematic equations instead of graphs?

Yes, absolutely! You can set up equations for the two phases (acceleration and deceleration) using $v = u + at$ and $s = ut + \frac{1}{2}at^2$. However, the graphical method using areas is often much faster and less prone to algebraic errors, as it provides a clear visual representation of the relative motion.

Question

On a large slippery ground, a boy left his dog sitting and walks away with a constant velocity $v_b = 2.0 	ext{ m/s}$. When he is $x_0 = 199 	ext{ m}$ away from the dog, the dog decides to catch him and thereafter move together. The dog cannot develop acceleration more than $a = 2.0 	ext{ m/s}^2$ in any direction. In what minimum time will the dog meet the boy?

Sigma Percentile · Accepted Answer

The problem of the boy and his dog is a classic kinematic chase that tests our ability to translate physical constraints into mathematical equations. While it can be solved using standard equations of motion, employing a velocity-time ($v-t$) graph transforms a tedious algebraic slog into an elegant geometric puzzle.

### The Anatomy of a Chase

Let's break down the physical reality. The boy is moving away at a constant velocity $v_b$. The dog, starting from rest, needs to catch him in the minimum possible time. However, there's a crucial constraint: the dog must "catch him and thereafter move together."

This means the dog cannot simply accelerate continuously and crash into the boy at high speed. It must accelerate to close the gap, but then it must hit the brakes (decelerate) so that at the exact moment it reaches the boy, its velocity perfectly matches the boy's velocity $v_b$. To minimize the total time, the dog must use its maximum possible acceleration $+a$ and its maximum possible deceleration $-a$.

### Visualizing the Motion: The $v-t$ Graph

Imagine plotting this on a $v-t$ graph. The boy's motion is simple: a horizontal line at $v = v_b$.

The dog's motion is more dynamic. It starts at the origin $(0,0)$, shoots upwards with a slope of $+a$ until it hits some peak velocity $v_{max}$, and then drops downwards with a slope of $-a$ until it intersects the boy's line at $v = v_b$. Because the magnitudes of acceleration and deceleration are equal, the dog's velocity profile forms a symmetric "tent" shape above the boy's constant velocity line.

### The Geometry of Relative Distance

Here is where the magic happens. We know that the area under a $v-t$ graph represents the distance traveled. Therefore, the area *between* the dog's curve and the boy's line represents the net relative distance the dog has gained on the boy.

Let's identify two key areas on our graph:
1.  **Area $A_2$ (The Orange Triangle):** In the very beginning, the dog's speed is less than the boy's speed. During this time, the boy is actually pulling further away! This area represents the distance the dog *loses*. It is a triangle with height $v_b$ and base $\frac{v_b}{a}$.
    $$A_2 = \frac{1}{2} \left(\frac{v_b}{a}ight) v_b = \frac{v_b^2}{2a}$$

2.  **Area $A_1$ (The Green Triangle):** Once the dog's speed exceeds $v_b$, it starts gaining ground. This area represents the distance the dog *gains*. It is a symmetric triangle with height $(v_{max} - v_b)$. Because the slopes are $+a$ and $-a$, its total base is $\frac{2(v_{max} - v_b)}{a}$.
    $$A_1 = \frac{1}{2} \left( \frac{2(v_{max} - v_b)}{a} ight) (v_{max} - v_b) = \frac{(v_{max} - v_b)^2}{a}$$

For the dog to catch the boy, the net distance gained must equal the initial separation $x_0$.
$$A_1 - A_2 = x_0$$

### The Master Equation

Substituting our geometric areas into the relative distance equation gives:
$$\frac{(v_{max} - v_b)^2}{a} - \frac{v_b^2}{2a} = x_0$$

We can easily solve this for the height of the green triangle:
$$(v_{max} - v_b)^2 = a x_0 + \frac{v_b^2}{2}$$
$$v_{max} - v_b = \sqrt{a x_0 + \frac{v_b^2}{2}}$$

The total time $t$ is simply the sum of the bases of our two triangles:
$$t = 	ext{Base of } A_2 + 	ext{Base of } A_1$$
$$t = \frac{v_b}{a} + \frac{2(v_{max} - v_b)}{a}$$

Substituting our expression for $(v_{max} - v_b)$ yields the master formula for the minimum time:
$$t = \frac{v_b}{a} + \frac{2}{a} \sqrt{a x_0 + \frac{v_b^2}{2}}$$
$$t = \frac{v_b}{a} + \sqrt{\frac{4a x_0 + 2v_b^2}{a^2}}$$

### The Final Sprint

With the heavy lifting done, we just need to plug in the numbers given in the problem: $v_b = 2.0 	ext{ m/s}$, $a = 2.0 	ext{ m/s}^2$, and $x_0 = 199 	ext{ m}$.

$$t = \frac{2.0}{2.0} + \sqrt{\frac{4(2.0)(199) + 2(2.0)^2}{(2.0)^2}}$$
$$t = 1 + \sqrt{\frac{1592 + 8}{4}}$$
$$t = 1 + \sqrt{\frac{1600}{4}}$$
$$t = 1 + \sqrt{400}$$
$$t = 1 + 20 = 21 	ext{ s}$$

The dog executes its perfect kinematic maneuver and catches the boy in exactly **21 seconds**.

Enter Numerical Value:

Visualized Solution

$v - t$ Graph Setup

Velocity Profile

Relative Distance

Calculating $A_{1}$ and $A_{2}$

Symmetry of $A_{1}$

Solving for $v_{ma x}$

Total Time $t$

Final Calculation

The Sigma Insight: Motion in a Straight Line

The Anatomy of a Chase

Visualizing the Motion

The Geometry of Relative Distance

The Master Equation

The Final Sprint

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