When you start your stopwatch, a particle moving on the x-axis is observed somewhere between the pos

Q: Why did we assume the velocity is constant?

In kinematics problems where no acceleration or force is mentioned, and the particle is simply 'observed' at various positions, the standard convention is to assume uniform motion (constant velocity) unless stated otherwise.

Q: Why does 'sometime during the fourth second' mean $3 < t < 4$?

The first second is from $t=0$ to $t=1$. The second is from $t=1$ to $t=2$. Following this pattern, the fourth second is the time interval strictly between $t=3$ and $t=4$.

Q: Why did we use $x(3) 22$ to find the maximum velocity?

Because the particle must cross $x=22$ *after* $t=3$. If it was moving any faster than the bound given by $x(3) < 22$, it would have crossed the 22m mark before the fourth second even began!

Q: How did the range of $x_0$ shrink from $(10, 12)$ to $(10, 11)$?

The constraints on velocity at $t=3$ and $t=12$ conflict with each other if $x_0$ is between 11 and 12. For the particle to satisfy both future position constraints simultaneously, it must have started strictly below $x=11$.

Q: Why did we substitute $x_0 = 10$ to find the minimum time, shouldn't a smaller distance take less time?

Time depends on both distance and velocity. To minimize time, we need the maximum possible velocity. The maximum velocity function $v_{max} = (22 - x_0)/3$ is largest when $x_0$ is smallest. The increase in velocity outweighs the slight increase in distance.

Question

When you start your stopwatch, a particle moving on the $x$-axis is observed somewhere between the positions $x = 10$ m and $x = 12$ m. Sometime during the fourth second, it passes the position $x = 22$ m and at the instant $t = 12$ s it is observed somewhere between the positions $x = 55$ m and $x = 60$ m. When do you expect its arrival at the position $x = 88$ m?

Sigma Percentile · Accepted Answer

### Mastering Kinematic Inequalities: The Bounding Box Method Imagine a particle moving along the x-axis. The question doesn't give us a neat, single equation for its motion. Instead, it gives us a series of observations—windows in time and space where the particle was spotted. Our mission is to predict when it will arrive at a future destination, $x = 88$ m. Because no acceleration is mentioned and the particle is simply "observed" at various positions, the standard convention is to assume uniform motion. Let's define the position of the particle as $x(t) = x_0 + vt$, where $x_0$ is the initial position and $v$ is the constant velocity. ### Translating Observations into Mathematics We are given three distinct constraints. First, at $t = 0$, the particle is between $10$ and $12$ meters. Mathematically, this is $10 < x_0 < 12$. Second, during the fourth second (which means strictly between $t = 3$ and $t = 4$), it crosses the $22$ meter mark. This implies that at exactly $t = 3$, it hasn't reached $22$ m yet, and by $t = 4$, it has passed it. This gives us $x(3) < 22$ and $x(4) > 22$. Finally, at $t = 12$ s, it is observed between $55$ and $60$ meters. This translates to $55 < x(12) < 60$. ### Extracting the Velocity Limits Now, let's extract the limits on the velocity $v$ based on our initial position $x_0$. From the condition $x(3) < 22$, we can write: $$x_0 + 3v < 22 \implies v < \frac{22 - x_0}{3}$$ This gives us an upper limit on the velocity. Similarly, from the condition $x(12) > 55$, we write: $$x_0 + 12v > 55 \implies v > \frac{55 - x_0}{12}$$ This provides a lower limit. Notice how, for any given initial position $x_0$, the velocity is trapped within a specific range. ### The Hidden Constraint on Initial Position There is a catch here. For a valid velocity to even exist, the upper limit must always be strictly greater than the lower limit. We must set these two limits against each other: $$\frac{55 - x_0}{12} < \frac{22 - x_0}{3}$$ By cross-multiplying, we get $55 - x_0 < 4(22 - x_0)$. Simplifying this reveals that $3x_0 < 33$, which means $x_0 < 11$. But wait! We already knew that $x_0 > 10$. So, our new, more accurate range for the initial position is strictly between $10$ and $11$ meters. This step is absolutely crucial; without it, we would calculate incorrect extreme times. ### Calculating the Extreme Arrival Times Now for the main question. When will the particle reach $x = 88$ m? The formula for time is $t = \frac{88 - x_0}{v}$. We need to find the minimum and maximum possible times. For the **minimum time** ($t_{min}$), the particle needs to travel as fast as possible. We substitute the maximum value of velocity, $v_{max} o \frac{22 - x_0}{3}$: $$t_{min} o \frac{88 - x_0}{(22 - x_0)/3} = 3 \left( 1 + \frac{66}{22 - x_0} ight)$$ To minimize this function, we need to substitute the smallest possible value of $x_0$, which is $10$. Upon calculating, we find $t_{min} o 3(1 + 5.5) = 19.5$ s. For the **maximum time** ($t_{max}$), the particle needs to travel as slowly as possible. We substitute the minimum value of velocity, $v_{min} o \frac{55 - x_0}{12}$: $$t_{max} o \frac{88 - x_0}{(55 - x_0)/12} = 12 \left( 1 + \frac{33}{55 - x_0} ight)$$ To maximize this, we substitute the largest possible value of $x_0$, which is $11$. Solving this gives $t_{max} o 12(1 + 0.75) = 21$ s. We can confidently conclude that the particle will arrive at the $88$ meter mark strictly between **$19.5$ s and $21$ s**. This brilliant problem beautifully demonstrates how multiple loose constraints can tightly bound a physical trajectory.

Visualized Solution

Mathematical Translation of Constraints

Extracting Velocity Limits

Intersection of Constraints

Refining the Initial Position

Setting Up the Time Function

Calculating Minimum Time

Calculating Maximum Time

Final Conclusion

The Sigma Insight: Motion in a Straight Line

Mastering Kinematic Inequalities

Translating Observations into Mathematics

Extracting the Velocity Limits

The Hidden Constraint on Initial Position

Calculating the Extreme Arrival Times

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