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 $a$ denotes the instantaneous acceleration of the particle, then:

• A. $a$ cannot remain positive for all $t$ in the interval $0 \leq t \leq 1
• B. $|a|$ cannot exceed 2 at any point in its path
• C. $|a|$ must be $\geq 4$ at some point of points in its path
• D. $a$ must change sign during the motion, but no other assertion can be made with the information given

Answer 1

Answer: (A), (C)

Options A and C are correct.

Answer 2

Let $x(t)$, $v(t)$, and $a(t)$ be the position, velocity, and acceleration of the particle at time $t$.

We are given the following conditions:

1. $x(0) = 0$
2. $v(0) = 0$ (starts from rest)
3. $x(1) = 1$
4. $v(1) = 0$ (comes to rest)

The motion is in the interval $0 \leq t \leq 1$.

First, let's consider the average velocity:

$\bar{v} = \frac{x(1) - x(0)}{1 - 0} = \frac{1 - 0}{1} = 1$.

By the Mean Value Theorem for derivatives, there exists $t_0 \in (0, 1)$ such that $v(t_0) = \bar{v} = 1$.

Now consider the velocity function $v(t)$. We have $v(0) = 0$, $v(1) = 0$, and $v(t_0) = 1$ for some $t_0 \in (0, 1)$.

Since $v(0) = 0$ and $v(t_0) = 1$, the velocity increases from 0 to 1 in the interval $[0, t_0]$.

Since $v(t_0) = 1$ and $v(1) = 0$, the velocity decreases from 1 to 0 in the interval $[t_0, 1]$.

Consider Option A: $a$ cannot remain positive for all $t$ in the interval $0 \leq t \leq 1$.

If $a(t) > 0$ for all $t \in [0, 1]$, then $v(t) = v(0) + \int_0^t a(\tau) d\tau = \int_0^t a(\tau) d\tau$. Since $a(\tau) > 0$ for $\tau > 0$, $v(t) > 0$ for $t \in (0, 1]$. This contradicts $v(1) = 0$.

If $a(t) \geq 0$ for all $t \in [0, 1]$, then $v(t)$ is non-decreasing. Since $v(0)=0$, $v(t) \geq 0$ for all $t \in [0, 1]$. If $a(t)$ is not identically zero on $[0, 1]$, then $v(t)$ would be strictly increasing over some interval, leading to $v(1) > 0$. If $a(t) = 0$ for all $t$, then $v(t) = 0$ for all $t$, which means $x(t) = x(0) = 0$ for all $t$, contradicting $x(1)=1$. Thus, $a(t)$ cannot be $\geq 0$ for all $t \in [0, 1]$. Similarly, $a(t)$ cannot be $\leq 0$ for all $t \in [0, 1]$ (this would lead to $v(t) \leq 0$ and $v(1) < 0$ unless $a(t)=0$ everywhere).

Therefore, $a(t)$ must take both positive and negative values in the interval $(0, 1)$. This means $a$ must change sign.

Option A is correct.

Consider Option D: $a$ must change sign during the motion, but no other assertion can be made with the information given.

We have shown that $a$ must change sign. However, we can make other assertions, as we will see by analyzing the magnitude of $a$. So, the second part of Option D is incorrect.

Consider the acceleration $a(t) = v'(t)$.

By the Mean Value Theorem applied to $v(t)$ on the interval $[0, t_0]$, there exists $t_1 \in (0, t_0)$ such that $a(t_1) = v'(t_1) = \frac{v(t_0) - v(0)}{t_0 - 0} = \frac{1 - 0}{t_0} = \frac{1}{t_0}$.

Since $t_0 \in (0, 1)$, $0 < t_0 < 1$, so $a(t_1) = \frac{1}{t_0} > 1$.

By the Mean Value Theorem applied to $v(t)$ on the interval $[t_0, 1]$, there exists $t_2 \in (t_0, 1)$ such that $a(t_2) = v'(t_2) = \frac{v(1) - v(t_0)}{1 - t_0} = \frac{0 - 1}{1 - t_0} = \frac{-1}{1 - t_0}$.

Since $t_0 \in (0, 1)$, $0 < 1 - t_0 < 1$, so $a(t_2) = \frac{-1}{1 - t_0} < -1$.

Thus, $|a(t_1)| = \frac{1}{t_0}$ and $|a(t_2)| = \frac{1}{1 - t_0}$.

Since $t_0 \in (0, 1)$, we have $\frac{1}{t_0} > 1$ and $\frac{1}{1 - t_0} > 1$.

At least one of these values must be $\geq 2$. If $\frac{1}{t_0} < 2$ and $\frac{1}{1 - t_0} < 2$, then $t_0 > 1/2$ and $1 - t_0 > 1/2$, so $t_0 < 1/2$. This is a contradiction.

Thus, $\max(\frac{1}{t_0}, \frac{1}{1 - t_0}) \geq 2$.

So, there exists a point where $|a| \geq 2$. This means Option B: $|a|$ cannot exceed 2 at any point in its path, is incorrect.

Now consider the minimum value that $\max(|a(t_1)|, |a(t_2)|)$ can take. This minimum occurs when $\frac{1}{t_0} = \frac{1}{1 - t_0}$, which implies $t_0 = 1/2$. In this case, $|a(t_1)| = 2$ and $|a(t_2)| = 2$. So, there are points where $|a| \geq 2$.

Let $M = \sup_{t \in [0, 1]} |a(t)|$. We have $M \geq |a(t_1)| = 1/t_0$ and $M \geq |a(t_2)| = 1/(1-t_0)$.

So, $M \geq \max(\frac{1}{t_0}, \frac{1}{1-t_0})$. Since this holds for any $t_0$ such that $v(t_0)=1$, and we know such a $t_0$ exists, $M \geq \min_{t_0 \in (0, 1)} \max(\frac{1}{t_0}, \frac{1}{1-t_0})$. The minimum value is 2, occurring at $t_0 = 1/2$.

So, the maximum magnitude of acceleration must be at least 2.

Let's try to get a stronger bound using integration.

We have $v(t) = \int_0^t a(\tau) d\tau$.

$x(1) - x(0) = \int_0^1 v(t) dt = 1$.

Also, $v(1) - v(0) = \int_0^1 a(t) dt = 0$.

Consider the integral $\int_0^1 t a(t) dt$. Using integration by parts:

$\int_0^1 t a(t) dt = \int_0^1 t v'(t) dt = [t v(t)]_0^1 - \int_0^1 v(t) dt = (1 \cdot v(1) - 0 \cdot v(0)) - \int_0^1 v(t) dt = (1 \cdot 0 - 0 \cdot 0) - 1 = -1$.

So, $\int_0^1 t a(t) dt = -1$.

Similarly, consider the integral $\int_0^1 (1-t) a(t) dt$. Using integration by parts:

$\int_0^1 (1-t) a(t) dt = \int_0^1 (1-t) v'(t) dt = [(1-t) v(t)]_0^1 - \int_0^1 (-1) v(t) d\tau = (0 \cdot v(1) - 1 \cdot v(0)) + \int_0^1 v(t) dt = (0 - 0) + 1 = 1$.

So, $\int_0^1 (1-t) a(t) dt = 1$.

Let $M = \sup_{t \in [0, 1]} |a(t)|$.

From $\int_0^1 t a(t) dt = -1$, we have $|\int_0^1 t a(t) dt| = 1$.

$|\int_0^1 t a(t) dt| \leq \int_0^1 |t a(t)| dt = \int_0^1 t |a(t)| dt \leq \int_0^1 t M dt = M \int_0^1 t dt = M [\frac{t^2}{2}]_0^1 = M \cdot \frac{1}{2}$.

So, $1 \leq M/2$, which implies $M \geq 2$.

From $\int_0^1 (1-t) a(t) dt = 1$, we have $|\int_0^1 (1-t) a(t) dt| = 1$.

$|\int_0^1 (1-t) a(t) dt| \leq \int_0^1 |(1-t) a(t)| dt = \int_0^1 (1-t) |a(t)| dt \leq \int_0^1 (1-t) M dt = M \int_0^1 (1-t) dt = M [t - \frac{t^2}{2}]_0^1 = M (1 - \frac{1}{2}) = M \cdot \frac{1}{2}$.

So, $1 \leq M/2$, which implies $M \geq 2$.

This method only guarantees $M \geq 2$.

Let's consider the integral $\int_0^1 (t - 1/2) a(t) dt$.

$\int_0^1 (t - 1/2) a(t) dt = \int_0^1 t a(t) dt - \frac{1}{2} \int_0^1 a(t) dt = -1 - \frac{1}{2}(0) = -1$.

So, $|\int_0^1 (t - 1/2) a(t) dt| = 1$.

$|\int_0^1 (t - 1/2) a(t) dt| \leq \int_0^1 |t - 1/2| |a(t)| dt \leq \int_0^1 |t - 1/2| M dt = M \int_0^1 |t - 1/2| dt$.

$\int_0^1 |t - 1/2| dt = \int_0^{1/2} -(t - 1/2) dt + \int_{1/2}^1 (t - 1/2) dt = \int_0^{1/2} (1/2 - t) dt + \int_{1/2}^1 (t - 1/2) dt$

$= [\frac{t}{2} - \frac{t^2}{2}]_0^{1/2} + [\frac{t^2}{2} - \frac{t}{2}]_{1/2}^1 = (\frac{1}{4} - \frac{1}{8}) - (0 - 0) + (\frac{1}{2} - \frac{1}{2}) - (\frac{1}{8} - \frac{1}{4}) = \frac{1}{8} + (0) - (-\frac{1}{8}) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$.

So, $1 \leq M \cdot \frac{1}{4}$, which implies $M \geq 4$.

Thus, the maximum magnitude of acceleration $|a|$ must be at least 4 at some point in the path.

This means $|a|$ must be $\geq 4$ at some point or points in its path.

Option C: $|a|$ must be $\geq 4$ at some point of points in its path, is correct.

Option B: $|a|$ cannot exceed 2 at any point in its path. This is incorrect as $M \geq 4$.

Let's summarize:

Option A is correct because $a$ must change sign.

Option B is incorrect because $|a|$ must be at least 4 at some point.

Option C is correct because $|a|$ must be at least 4 at some point.

Option D is incorrect because we can make other assertions (like $|a| \geq 4$ at some point).

Since the question asks for the correct option(s), and this is a multiple choice question, we should check if multiple options can be correct. Based on our analysis, A and C are correct. However, in JEE/NEET, usually only one option is correct unless explicitly stated otherwise. Let's re-read the question and options carefully. The wording suggests that we need to choose one or more assertions that are true.

Let's consider if there's a possibility that only one option is intended to be correct. Option A states that $a$ cannot remain positive for all $t$. This is true and implies $a$ must change sign (or be zero everywhere, which is ruled out). Option C states that $|a|$ must be $\geq 4$ at some point. This is a stronger statement about the magnitude of $a$. Option D says $a$ must change sign, but no other assertion can be made. This is false because we could assert $|a| \geq 4$ at some point.

If this were a single-choice question, there might be an issue with the options provided, as both A and C seem correct. However, Option C provides a more specific and stronger conclusion about the motion, which is derived from the given conditions. Option A is a consequence of the velocity starting and ending at zero while displacement is non-zero. Option C is a consequence of the specific values of displacement and time interval.

If forced to choose a single option, Option C is a stronger statement than Option A.

Final Answer: The final answer is $\boxed{A, C}$