A particle moving continuously in the positive x-direction passes the positions x = 9 m and 17 m at

Q: Why can't we assume the particle is moving with uniform speed?

The problem only states it moves 'continuously in the positive x-direction'. It never says the speed is constant. A particle can speed up and slow down while still having the same average velocity over a specific interval.

Q: If $v_{\text{avg}}$ is $4\text{ m/s}$, doesn't that mean $x(0)$ must be $5\text{ m}$ to reach $9\text{ m}$ at $t=1\text{ s}$?

That would only be true if the velocity was strictly uniform ($4\text{ m/s}$ at every instant). Since it could be accelerating or decelerating, it might have covered $2\text{ m}$ in the first second and $6\text{ m}$ in the next, for example.

Q: What does 'moving continuously in the positive x-direction' mean mathematically?

It means the velocity $v(t)$ is always greater than or equal to zero. The particle never turns back. However, the magnitude of $v(t)$ can still fluctuate.

Q: Can we find the average velocity for $[3\text{ s}, 6\text{ s}]$?

No. We know the total displacement for $[0, 6\text{ s}]$ is $24\text{ m}$, and for $[1, 3\text{ s}]$ it's $8\text{ m}$. But we don't know how the remaining $16\text{ m}$ of displacement is distributed between the $[0, 1\text{ s}]$ and $[3, 6\text{ s}]$ intervals.

Q: Is it possible to construct a motion equation that satisfies all conditions but has $x(0) \neq 5$?

Absolutely! For instance, imagine the particle moves very slowly at first, then speeds up. As long as $x(1)=9$, $x(3)=17$, and $x(6) - x(0) = 24$, the conditions are met.

Question

A particle moving continuously in the positive $x$-direction passes the positions $x = 9	ext{ m}$ and $17	ext{ m}$ at the instants $t = 1	ext{ s}$ and $3	ext{ s}$ respectively. Its average velocities in the time intervals $[1	ext{ s}, 3	ext{ s}]$ and $[0	ext{ s}, 6	ext{ s}]$ are equal. Which of the following statements is/are correct?

Options:

(A) It was at $x = 5	ext{ m}$ at $t = 0	ext{ s}$.

(B) It is moving with a uniform speed.

(C) Average velocity in the interval $[3	ext{ s}, 6	ext{ s}]$ is $4	ext{ m/s}$.

(D) Information is insufficient to decide.

Sigma Percentile · Accepted Answer

### The Illusion of Average Velocity

Imagine you are tracking a road trip. You know exactly where the car was at 1:00 PM and where it was at 3:00 PM. From this, you can easily calculate the average speed for that two-hour window. But does that tell you how fast the car was going at 1:30 PM? Or where it started at noon?

This problem plays on that exact intuition. It gives us a few concrete data points and tempts us to make sweeping assumptions about the entire journey. Let's break down the physics mathematically and see why we must tread carefully.

### Breaking Down the Knowns

We are given two specific snapshots of a particle moving continuously in the positive $x$-direction:
- At $t_1 = 1	ext{ s}$, the position is $x_1 = 9	ext{ m}$.
- At $t_2 = 3	ext{ s}$, the position is $x_2 = 17	ext{ m}$.

The fundamental tool we need here is the formula for average velocity, which is the total displacement divided by the total time taken:

$$v_{	ext{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$$

Let's apply this to the interval $[1	ext{ s}, 3	ext{ s}]$:

$$v_{	ext{avg}, 1	o3} = \frac{17 - 9}{3 - 1} = \frac{8}{2} = 4	ext{ m/s}$$

So, we know for an absolute fact that between the 1st and 3rd second, the particle covered ground at an average rate of $4	ext{ m/s}$.

### Expanding to the Full Journey

The problem provides a crucial piece of information: the average velocity for the entire interval from $t = 0	ext{ s}$ to $t = 6	ext{ s}$ is exactly the same as the average velocity we just calculated.

Therefore, $v_{	ext{avg}, 0	o6} = 4	ext{ m/s}$.

Let's plug this into our average velocity formula for the $[0, 6]$ interval:

$$v_{	ext{avg}, 0	o6} = \frac{x(6) - x(0)}{6 - 0} = 4$$

Multiplying both sides by 6, we get:

$$x(6) - x(0) = 24	ext{ m}$$

This equation tells us that the total displacement over the 6 seconds is $24	ext{ m}$.

### The Trap of Uniform Motion

Here is where many students make a fatal error. They look at the average velocity of $4	ext{ m/s}$ and subconsciously assume the particle is moving with a **constant, uniform speed** of $4	ext{ m/s}$ at every single instant.

If the speed were uniform, the position function would be $x(t) = 4t + 5$. 
Let's test this:
- $x(0) = 4(0) + 5 = 5	ext{ m}$
- $x(1) = 4(1) + 5 = 9	ext{ m}$ (Matches given data!)
- $x(3) = 4(3) + 5 = 17	ext{ m}$ (Matches given data!)
- $x(6) = 4(6) + 5 = 29	ext{ m}$

If this were the *only* possible motion, then option (a) "It was at $x = 5	ext{ m}$ at $t = 0	ext{ s}$" would be correct.

**But is uniform motion guaranteed?** No! The problem only states the particle moves "continuously in the positive $x$-direction". It could be accelerating, decelerating, or moving erratically as long as it never turns back.

Imagine a different scenario where the particle moves very slowly at first, and then speeds up. What if $x(0) = 6	ext{ m}$ and $x(6) = 30	ext{ m}$? 
The displacement is $30 - 6 = 24	ext{ m}$, which perfectly satisfies our equation $x(6) - x(0) = 24$. 
There is no physical law preventing the particle from being at $x = 6	ext{ m}$ at $t = 0$, reaching $x = 9	ext{ m}$ at $t = 1$, and so on.

### Final Conclusion

Because we have one equation ($x(6) - x(0) = 24$) but two unknown variables ($x(6)$ and $x(0)$), we have a system with infinitely many valid solutions.

We cannot definitively prove where the particle started, where it ended, or what its average velocity was in the specific $[3	ext{ s}, 6	ext{ s}]$ sub-interval. Therefore, the only logically sound conclusion is that the **information is insufficient to decide**.

Select Answer:

Visualized Solution

$t_{1} = 1 s, x_{1} = 9 m$

$v_{avg} = \frac{Δ x}{Δ t}$

$v_{avg, 1 \to 3} = \frac{17 - 9}{3 - 1}$

$v_{avg, 1 \to 3} = 4 m/s$

$v_{avg, 0 \to 6} = 4 m/s$

$x (6) - x (0) = 24 m$

\text{Analyzing Options}

\text{Conclusion}

The Sigma Insight: Motion in a Straight Line

The Illusion of Average Velocity

Breaking Down the Knowns

Expanding to the Full Journey

The Trap of Uniform Motion

Final Conclusion

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​=1 s,x1​=9 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vavg​=ΔtΔx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vavg,1→3​=3−117−9​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vavg,1→3​=4 m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vavg,0→6​=4 m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x(6)−x(0)=24 m