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:

Step 1: Define the boundary conditions.

The particle starts from rest at $x = 0$ at $t = 0$, and comes to rest at $x = 1$ at $t = 1$.

$$x(0) = 0, \quad v(0) = 0$$ $$x(1) = 1, \quad v(1) = 0$$

Step 2: Analyze the sign of the acceleration $\alpha(t)$.

The acceleration is the rate of change of velocity, $\alpha(t) = \dfrac{dv}{dt}$. Integrate the acceleration over the interval $[0, 1]$:

$$\int_{0}^{1} \alpha(t) \, dt = v(1) - v(0) = 0$$

Since the integral of $\alpha(t)$ is zero and the particle must move (so $\alpha(t)$ is not identically zero), $\alpha(t)$ must take both positive and negative values. Therefore, $\alpha(t)$ cannot remain positive at all times.

Step 3: Establish a bound on the maximum acceleration magnitude.

Let $A$ be the maximum value of the magnitude of acceleration, such that $|\alpha(t)| \le A$ for all $t \in [0, 1]$.

Using the initial condition $v(0) = 0$:

$$|v(t)| = \left| \int_{0}^{t} \alpha(u) \, du \right| \le \int_{0}^{t} |\alpha(u)| \, du \le At$$

Using the final condition $v(1) = 0$:

$$|v(t)| = \left| -\int_{t}^{1} \alpha(u) \, du \right| \le \int_{t}^{1} |\alpha(u)| \, du \le A(1 - t)$$

Combining these two inequalities yields:

$$|v(t)| \le A \min(t, 1 - t)$$

Step 4: Integrate the velocity bound to find the minimum value of $A$.

The total displacement is the integral of the velocity:

$$\int_{0}^{1} v(t) \, dt = x(1) - x(0) = 1$$

Using the inequality for $|v(t)|$:

$$1 = \int_{0}^{1} v(t) \, dt \le \int_{0}^{1} |v(t)| \, dt \le \int_{0}^{1} A \min(t, 1 - t) \, dt$$

Evaluate the integral of the minimum function:

$$\int_{0}^{1} \min(t, 1 - t) \, dt = \int_{0}^{1/2} t \, dt + \int_{1/2}^{1} (1 - t) \, dt = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$$

Substitute this back into the inequality:

$$1 \le A \cdot \frac{1}{4} \implies A \ge 4$$

Thus, the acceleration magnitude $|\alpha(t)|$ must be at least $4$ at some point in the interval $0 < t < 1$.

Answer: The correct assertions are:

• $\alpha$ cannot remain positive at all times (it must change sign).
• $|\alpha| \ge 4$ at some point in the interval $0 < t < 1$.