Consider a body of mass 1.0 kg at rest at th | vedclass

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(A) The force applied on the body is $\overrightarrow{F}=(\alpha t) \hat{i}+\beta \hat{j}$.Given $\alpha=1.0 \ Ns^{-1}$ and $\beta=1.0 \ N$,we have $\

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$A, C$

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(A) The force applied on the body is $\overrightarrow{F}=(\alpha t) \hat{i}+\beta \hat{j}$.Given $\alpha=1.0 \ Ns^{-1}$ and $\beta=1.0 \ N$,we have $\overrightarrow{F}=t \hat{i}+\hat{j}$.Using Newton's second law,$m \frac{d\overrightarrow{v}}{dt} = \overrightarrow{F}$,with $m=1.0 \ kg$,we get $\frac{d\overrightarrow{v}}{dt} = t \hat{i}+\hat{j}$.Integrating with respect to time $t$ (with $\overrightarrow{v}=0$ at $t=0$),we get $\overrightarrow{v} = \int_{0}^{t} (t \hat{i}+\hat{j}) dt = \frac{t^2}{2} \hat{i} + t \hat{j}$.At $t=1 \ s$,$\overrightarrow{v} = \frac{1}{2} \hat{i} + \hat{j} = \frac{1}{2}(\hat{i} + 2\hat{j}) \ ms^{-1}$. Thus,statement $(C)$ is true.Integrating velocity to find position $\overrightarrow{r}$ (with $\overrightarrow{r}=0$ at $t=0$),we get $\overrightarrow{r} = \int_{0}^{t} (\frac{t^2}{2} \hat{i} + t \hat{j}) dt = \frac{t^3}{6} \hat{i} + \frac{t^2}{2} \hat{j}$.At $t=1 \ s$,$\overrightarrow{r} = \frac{1}{6} \hat{i} + \frac{1}{2} \hat{j}$.The displacement is $\overrightarrow{s} = \overrightarrow{r}(1) - \overrightarrow{r}(0) = \frac{1}{6} \hat{i} + \frac{1}{2} \hat{j}$.The magnitude of displacement is $|\overrightarrow{s}| = \sqrt{(\frac{1}{6})^2 + (\frac{1}{2})^2} = \sqrt{\frac{1}{36} + \frac{9}{36}} = \sqrt{\frac{10}{36}} = \frac{\sqrt{10}}{6} \ m$. Thus,statement $(D)$ is false.The torque is $\vec{	au} = \overrightarrow{r} 	imes \overrightarrow{F} = (\frac{t^3}{6} \hat{i} + \frac{t^2}{2} \hat{j}) 	imes (t \hat{i} + \hat{j})$.At $t=1 \ s$,$\vec{	au} = (\frac{1}{6} \hat{i} + \frac{1}{2} \hat{j}) 	imes (1 \hat{i} + 1 \hat{j}) = (\frac{1}{6} - \frac{1}{2}) \hat{k} = -\frac{1}{3} \hat{k} \ Nm$.The magnitude is $|\vec{	au}| = \frac{1}{3} \ Nm$. Thus,statement $(A)$ is true and $(B)$ is false.

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