A particle of mass m moves on the x-axis as follows : it starts from rest at t = 0 from the point x

Q: Wait, why do we reject the idea of constant acceleration?

If the acceleration were constant and positive, the particle's velocity would keep increasing, and it would never come back to rest at $t=1$. If it were constant and negative, it would move backwards. To start from rest and end at rest, it must accelerate to gain speed, and then decelerate (negative acceleration) to stop.

Q: Why can't we just use the standard equations of motion like $s = ut + \frac{1}{2}at^2$?

The standard equations of motion are strictly for *constant* acceleration. In this problem, because the particle starts at rest and ends at rest, the acceleration *must* change (it must be positive to speed up, then negative to slow down). Therefore, we cannot use the constant acceleration formulas for the entire journey. We must rely on calculus and graphical analysis!

Q: How did we come up with the magic number 4 for the acceleration bound?

It comes from optimizing the area! To get the maximum area (displacement) with the minimum possible peak acceleration, you want a symmetric triangular velocity graph. If you set the area to 1 and the base to 1, the peak velocity must be 2. To reach a velocity of 2 in 0.5 seconds, the acceleration must be $2 / 0.5 = 4$. Any deviation from this perfect symmetry requires an even higher acceleration to maintain the same area.

Q: What if the velocity graph is a smooth curve, like a parabola, instead of a sharp triangle?

Great question! Even if the graph is a smooth curve, it must still enclose an area of 1. If you try to draw a smooth curve that stays entirely below the $v=4t$ and $v=4(1-t)$ bounding lines, its area will always be strictly less than 1. To get the area up to 1, the curve *must* pierce through those bounding lines at some point, which mathematically guarantees that its slope (acceleration) will exceed 4.

Q: Does the particle have to move in a straight line without turning back?

The problem states it moves from $x=0$ to $x=1$. Even if it moved back and forth, the *net* displacement is 1. To cover a net displacement of 1 while starting and stopping at rest in 1 second, the peak velocity and thus the peak acceleration would actually have to be *even higher* than if it just moved straight. So the minimum bound of 4 still holds!

Question

A particle of mass $m$ moves on the $x$-axis as follows : it starts from rest at $t = 0$ from the point $x = 0$ and comes to rest at $t = 1$ at the point $x = 1$. No other information is available about its motion at intermediate times ($0 < t < 1$). If $\alpha$ denotes the instantaneous acceleration of the particle, then

Options:

(A) $\alpha$ cannot remain positive for all $t$ in the interval $0 \le t \le 1$

(B) $|\alpha|$ cannot exceed 2 at any point in its path

(C) $|\alpha|$ must be $\ge 4$ at some point or points in its path

(D) $\alpha$ must change sign during the motion, but no other assertion can be made with the information given

Sigma Percentile · Accepted Answer

### The Hidden Limits of Motion: Unveiling the Acceleration Trap

Imagine you are standing at the starting line of a 1-meter sprint. You start from absolute rest. The clock ticks. Exactly one second later, you cross the finish line and instantly freeze, coming to a complete halt.

This is the physical reality of the particle in our problem. It seems so incredibly simple, yet it mathematically forces the particle to undergo extreme physical stress. Let's uncover why this happens and how we can use graphical analysis to expose the hidden limits of its acceleration.

### Phase 1: The Inevitable Reversal

Let's think about the velocity of this particle. It starts at $v = 0$ and ends at $v = 0$. However, it must cover a distance of 1 meter in between.

To move forward, the particle must gain speed. This means its velocity must increase, which requires a **positive acceleration** ($\alpha > 0$). But it cannot keep speeding up forever; it has to stop at exactly $t = 1$ s. To bring the velocity back down to zero, the particle must hit the brakes. This requires a **negative acceleration** ($\alpha < 0$).

Therefore, the acceleration cannot possibly remain positive for the entire journey. It must change sign. This simple logical deduction immediately proves that **Option (a)** is correct.

### Phase 2: The Geometric Translation

To truly see the hidden constraints of this motion, we must translate our physical reality into geometry. We do this by plotting a velocity-time ($v-t$) graph.

In kinematics, the area under a $v-t$ graph represents the total displacement. We are given that the total displacement is exactly 1 meter, and the total time is exactly 1 second.

Mathematically, this means:
$$ \int_0^1 v(t) dt = 1 $$

Our goal is to find the minimum possible value for the maximum acceleration. How hard does the particle *have* to accelerate to cover this area in this time?

### Phase 3: The Speed Limit (A Proof by Contradiction)

Let's play a game of 'What If?'. Let's assume the opposite of what we want to prove. Let's assume that the magnitude of the acceleration is strictly less than 4 at all times ($|\alpha| < 4$).

If the acceleration is capped at a value less than 4, there is a strict "speed limit" on how fast the particle can gain velocity. Starting from $t=0$, the maximum possible velocity it could have at any time $t$ is bounded by the line $v = 4t$.

Similarly, looking backwards from the finish line at $t=1$, the particle must decelerate to zero. If the maximum deceleration is also capped at a magnitude less than 4, the velocity must be bounded by the line $v = 4(1-t)$.

These two lines, $v = 4t$ and $v = 4(1-t)$, form a massive bounding triangle. If our assumption ($|\alpha| < 4$) is true, the actual velocity curve of the particle must lie **strictly inside** this triangle.

### Phase 4: The Area Trap

Let's calculate the area of this ultimate bounding triangle.

The base of the triangle is the total time, which is 1 second. The peak of the triangle occurs where the two bounding lines intersect: $4t = 4(1-t)$, which gives $t = 0.5$ s. At this time, the peak velocity is $v = 4(0.5) = 2$ m/s.

The area of this bounding triangle is:
$$ 	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 2 = 1 	ext{ m} $$

Here is the trap! The area of the *maximum possible bounding triangle* is exactly 1.

If the particle's acceleration is strictly less than 4 everywhere, its velocity curve must lie *strictly inside* this triangle. And if a shape lies strictly inside another shape, its area must be strictly less. This means the particle's total displacement would be strictly less than 1 meter.

### The Inescapable Conclusion

But we know for a fact that the displacement is exactly 1 meter!

This is a mathematical contradiction. Our initial assumption must be wrong. To achieve an area of exactly 1, the velocity curve cannot stay trapped inside that bounding triangle. It must pierce through those bounding lines at some point.

When the curve pierces the bounding lines, its slope (which is the acceleration) must be steeper than the slope of the lines. Since the lines have a slope of $\pm 4$, the particle's acceleration must be greater than or equal to 4 at some point during the journey.

Thus, $|\alpha| \ge 4$ at some point or points in its path. **Option (c)** is brilliantly correct.

Select Answer:

Visualized Solution

Initial Conditions

Velocity Variation

Sign of Acceleration

The $v - t$ Graph

Area as Displacement

Proof by Contradiction

Forward Velocity Bound

Backward Velocity Bound

The Bounding Triangle

Triangle Dimensions

Maximum Possible Area

The Area Contradiction

The Final Conclusion

The Sigma Insight: Motion Graphs

The Hidden Limits of Motion

Phase 1

Phase 2

Phase 3

Phase 4

The Inescapable Conclusion

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