Step-by-step solution:

The problem provides the mass of a body, its initial conditions (at rest at the origin), and the force acting on it as a function of time. We need to evaluate four statements about its motion and the torque acting on it at $t=1.0s$.

Given:

• Mass of the body, $m = 1.0\,kg$
• Initial position, $\overrightarrow r (0) = \overrightarrow 0$
• Initial velocity, $\overrightarrow v (0) = \overrightarrow 0$
• Applied force, $\overrightarrow F = (\alpha t \widehat i + \beta \widehat j)$, with $\alpha = 1.0\,Ns^{-1}$ and $\beta = 1.0\,N$.
• So, $\overrightarrow F(t) = (t \widehat i + \widehat j)\,N$.

1. Find the acceleration of the body: Using Newton's second law, $\overrightarrow F = m\overrightarrow a$. $\overrightarrow a(t) = \frac{\overrightarrow F(t)}{m} = \frac{(t \widehat i + \widehat j)}{1.0} = (t \widehat i + \widehat j)\,m/s^2$

2. Find the velocity of the body: Velocity is the integral of acceleration with respect to time. We use the initial condition $\overrightarrow v(0) = \overrightarrow 0$. $\overrightarrow v(t) = \overrightarrow v(0) + \int_0^t \overrightarrow a(t')\,dt'$ $\overrightarrow v(t) = \overrightarrow 0 + \int_0^t (t' \widehat i + \widehat j)\,dt'$ $\overrightarrow v(t) = \left[ \frac{t'^2}{2} \widehat i + t' \widehat j \right]_0^t$ $\overrightarrow v(t) = \left( \frac{t^2}{2} \widehat i + t \widehat j \right)\,m/s$

3. Find the position of the body: Position is the integral of velocity with respect to time. We use the initial condition $\overrightarrow r(0) = \overrightarrow 0$. $\overrightarrow r(t) = \overrightarrow r(0) + \int_0^t \overrightarrow v(t')\,dt'$ $\overrightarrow r(t) = \overrightarrow 0 + \int_0^t \left( \frac{t'^2}{2} \widehat i + t' \widehat j \right)\,dt'$ $\overrightarrow r(t) = \left[ \frac{t'^3}{6} \widehat i + \frac{t'^2}{2} \widehat j \right]_0^t$ $\overrightarrow r(t) = \left( \frac{t^3}{6} \widehat i + \frac{t^2}{2} \widehat j \right)\,m$

4. Calculate quantities at $t = 1.0s$ and evaluate the options:

At $t = 1.0s$:

• Position vector: $\overrightarrow r(1) = \left( \frac{1^3}{6} \widehat i + \frac{1^2}{2} \widehat j \right) = \left( \frac{1}{6} \widehat i + \frac{1}{2} \widehat j \right)\,m$
• Force vector: $\overrightarrow F(1) = (1 \widehat i + \widehat j)\,N$

Evaluation of Option A and B: The torque $\overrightarrow \tau$ about the origin is given by $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$. At $t=1.0s$: $\overrightarrow \tau(1) = \overrightarrow r(1) \times \overrightarrow F(1) = \left( \frac{1}{6} \widehat i + \frac{1}{2} \widehat j \right) \times (\widehat i + \widehat j)$ $= \frac{1}{6}(\widehat i \times \widehat i) + \frac{1}{6}(\widehat i \times \widehat j) + \frac{1}{2}(\widehat j \times \widehat i) + \frac{1}{2}(\widehat j \times \widehat j)$ Since $\widehat i \times \widehat i = 0$, $\widehat j \times \widehat j = 0$, $\widehat i \times \widehat j = \widehat k$, and $\widehat j \times \widehat i = -\widehat k$: $\overrightarrow \tau(1) = 0 + \frac{1}{6}\widehat k + \frac{1}{2}(-\widehat k) + 0 = \left( \frac{1}{6} - \frac{1}{2} \right)\widehat k = \left( \frac{1 - 3}{6} \right)\widehat k = -\frac{2}{6}\widehat k = -\frac{1}{3}\widehat k\,Nm$

• Option A: The magnitude of the torque is $|\overrightarrow \tau(1)| = | -\frac{1}{3} | = \frac{1}{3}\,Nm$. This statement is true.
• Option B: The direction of the torque is along the $-\widehat k$ vector, not the $+\widehat k$ vector. This statement is false.

Evaluation of Option C: At $t = 1.0s$, the velocity is: $\overrightarrow v(1) = \left( \frac{1^2}{2} \widehat i + 1 \widehat j \right) = \left( \frac{1}{2} \widehat i + \widehat j \right)\,m/s$ Factoring out $\frac{1}{2}$: $\overrightarrow v(1) = \frac{1}{2} (\widehat i + 2\widehat j)\,m/s$

• Option C: The velocity of the body at $t=1s$ is $\overrightarrow v = \frac{1}{2}(\widehat i + 2\widehat j)m{s^{ - 1}}$. This statement is true.

Evaluation of Option D: The displacement of the body at $t=1s$ is $\Delta \overrightarrow r = \overrightarrow r(1) - \overrightarrow r(0) = \overrightarrow r(1)$. The magnitude of the displacement is $|\overrightarrow r(1)|$. $|\overrightarrow r(1)| = \left| \frac{1}{6} \widehat i + \frac{1}{2} \widehat j \right| = \sqrt{\left(\frac{1}{6}\right)^2 + \left(\frac{1}{2}\right)^2}$ $= \sqrt{\frac{1}{36} + \frac{1}{4}} = \sqrt{\frac{1+9}{36}} = \sqrt{\frac{10}{36}} = \frac{\sqrt{10}}{6}\,m$

• Option D: The magnitude of displacement is $\frac{1}{6}m$. This is incorrect, as it is $\frac{\sqrt{10}}{6}m$. This statement is false.

Conclusion: The true statements are A and C.