A material point moving along a straight line enters an 87.5 m segment with speed 5.0 m/s and leaves

Q: Why does the particle have to accelerate at the maximum rate to minimize time?

To cover a fixed distance in the shortest time, the average speed must be as high as possible. Accelerating at the maximum rate right from the start ensures the particle reaches its top speed quickly and spends more time traveling at that higher speed.

Q: Can the particle accelerate and then decelerate?

The problem states the acceleration is 'unidirectional,' meaning it doesn't change sign. Since the final speed (10 m/s) is greater than the initial speed (5 m/s), the acceleration must always be positive or zero. It cannot decelerate.

Q: Why can't the particle just travel at 10 m/s the whole way?

The particle enters the segment at 5 m/s. It takes time and distance to accelerate to 10 m/s. It cannot instantaneously jump to 10 m/s because the maximum acceleration is limited to 1.0 m/s².

Q: How do we know the particle doesn't exceed 10 m/s in the middle of the segment?

If it exceeded 10 m/s, it would have to decelerate to reach 10 m/s at the exit. But the acceleration is unidirectional, so deceleration is not allowed. Therefore, 10 m/s is the maximum speed it can achieve.

Q: What if the maximum acceleration was higher, say 5 m/s²?

A higher maximum acceleration would allow the particle to reach 10 m/s in a shorter distance and time. This would decrease the minimum time and increase the maximum average acceleration. The maximum time would also decrease slightly.

Question

A material point moving along a straight line enters an $87.5 	ext{ m}$ segment with speed $5.0 	ext{ m/s}$ and leaves with speed $10 	ext{ m/s}$. The particle crosses the segment with unidirectional acceleration that never exceeds $1.0 	ext{ m/s}^2$. Find range of average acceleration of the point on this segment?

Sigma Percentile · Accepted Answer

### The Core Concept: Time is the Key

First, let's look at the definition of average acceleration. It is simply the total change in velocity divided by the total time taken:
$$a_{	ext{av}} = \frac{\Delta v}{\Delta t} = \frac{10 - 5}{t} = \frac{5}{t}$$

Notice something interesting? The numerator is fixed at $5 	ext{ m/s}$. The only variable that can change is the total time, $t$. This means that to find the range of average acceleration, we don't need to worry about complex acceleration functions; we just need to find the **minimum** and **maximum** possible times it takes to cross the $87.5 	ext{ m}$ segment.

Because $a_{	ext{av}}$ is inversely proportional to $t$, the minimum time will give us the maximum average acceleration, and the maximum time will give us the minimum average acceleration.

### The Need for Speed: Minimizing Time

How do we cross a fixed distance in the shortest possible time? We need to travel as fast as possible!

To maximize our speed throughout the journey, the particle should accelerate at its maximum allowed rate ($1.0 	ext{ m/s}^2$) right from the starting line. Once it hits the target exit speed of $10 	ext{ m/s}$, it should stop accelerating and just cruise at that top speed for the rest of the segment.

Let's calculate the distance and time for this initial "sprint". Using the third equation of motion ($v^2 - u^2 = 2as$), the distance covered while accelerating is:
$$s_1 = \frac{10^2 - 5^2}{2(1.0)} = 37.5 	ext{ m}$$

The time taken for this sprint is:
$$t_1 = \frac{10 - 5}{1.0} = 5 	ext{ s}$$

Out of the total $87.5 	ext{ m}$, we have covered $37.5 	ext{ m}$. The remaining distance is $87.5 - 37.5 = 50 	ext{ m}$. The particle travels this remaining distance at a constant speed of $10 	ext{ m/s}$. The time for this cruising phase is:
$$t_2 = \frac{50}{10} = 5 	ext{ s}$$

So, the absolute minimum time to cross the segment is $t_{	ext{min}} = 5 + 5 = 10 	ext{ s}$.

### Taking the Scenic Route: Maximizing Time

Now, let's flip the scenario. What if we want to take the longest possible time? We should travel as slowly as possible!

To minimize our speed throughout the journey, the particle should just lazily coast at its initial speed of $5 	ext{ m/s}$ for as long as it possibly can. It should only start accelerating at the very last moment, at the maximum rate, so that it hits exactly $10 	ext{ m/s}$ just as it crosses the finish line.

The acceleration phase at the end is identical to our previous calculation: it takes $37.5 	ext{ m}$ and $5 	ext{ s}$ to ramp up from $5 	ext{ m/s}$ to $10 	ext{ m/s}$.

This leaves $50 	ext{ m}$ to be covered at the slow initial speed of $5 	ext{ m/s}$. The time for this slow phase is:
$$t_2 = \frac{50}{5} = 10 	ext{ s}$$

Adding them up, the absolute maximum time to cross the segment is $t_{	ext{max}} = 10 + 5 = 15 	ext{ s}$.

### The Final Calculation

We have our time bounds! The time $t$ must be between $10 	ext{ s}$ and $15 	ext{ s}$. Now we just plug these extreme values back into our average acceleration formula.

The maximum average acceleration occurs when the time is minimized:
$$a_{	ext{av, max}} = \frac{5}{10} = 0.5 	ext{ m/s}^2$$

The minimum average acceleration occurs when the time is maximized:
$$a_{	ext{av, min}} = \frac{5}{15} = \frac{1}{3} \approx 0.33 	ext{ m/s}^2$$

Therefore, the range of the average acceleration is:
**$$0.33 	ext{ m/s}^2 \le a_{	ext{av}} \le 0.5 	ext{ m/s}^2$$**

By visualizing the extremes of motion—the fastest possible sprint and the slowest possible crawl—we elegantly bounded the average acceleration without needing any complex calculus. This is the power of physical intuition!

Visualized Solution

Understanding the Goal

The Velocity-Time Graph

Minimizing Time (Maximum Average Acceleration)

Calculating Minimum Time

Maximizing Time (Minimum Average Acceleration)

Calculating Maximum Time

Range of Average Acceleration

The Sigma Insight: Motion in a Straight Line

The Core Concept

The Need for Speed

Taking the Scenic Route

The Final Calculation

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