The position vector vec{r} of a particle of m | vedclass

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(C) The position vector is $\vec{r}(t) = \alpha t^3 \hat{i} + \beta t^2 \hat{j}$.Velocity $\vec{v}(t) = \frac{d\vec{r}}{dt} = 3\alpha t^2 \hat{i} + 2\

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$A, B, D$

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(C) The position vector is $\vec{r}(t) = \alpha t^3 \hat{i} + \beta t^2 \hat{j}$.Velocity $\vec{v}(t) = \frac{d\vec{r}}{dt} = 3\alpha t^2 \hat{i} + 2\beta t \hat{j}$.At $t = 1 \ s$,$\vec{v} = 3(10/3)(1)^2 \hat{i} + 2(5)(1) \hat{j} = (10 \hat{i} + 10 \hat{j}) \ m \ s^{-1}$. Thus,$(A)$ is correct.Position at $t = 1 \ s$: $\vec{r} = (10/3) \hat{i} + 5 \hat{j}$.Angular momentum $\vec{L} = m(\vec{r} 	imes \vec{v}) = 0.1 [((10/3) \hat{i} + 5 \hat{j}) 	imes (10 \hat{i} + 10 \hat{j})] = 0.1 [(100/3) \hat{k} - 50 \hat{k}] = 0.1 [(-50/3) \hat{k}] = -(5/3) \hat{k} \ N \ m \ s$. Thus,$(B)$ is correct.Acceleration $\vec{a}(t) = \frac{d\vec{v}}{dt} = 6\alpha t \hat{i} + 2\beta \hat{j}$.At $t = 1 \ s$,$\vec{a} = 6(10/3)(1) \hat{i} + 2(5) \hat{j} = 20 \hat{i} + 10 \hat{j}$.Force $\vec{F} = m\vec{a} = 0.1(20 \hat{i} + 10 \hat{j}) = (2 \hat{i} + 1 \hat{j}) \ N$. Thus,$(C)$ is correct.Torque $\vec{	au} = \vec{r} 	imes \vec{F} = ((10/3) \hat{i} + 5 \hat{j}) 	imes (2 \hat{i} + 1 \hat{j}) = (10/3) \hat{k} - 10 \hat{k} = -(20/3) \hat{k} \ N \ m$. Thus,$(D)$ is correct.Therefore,statements $(A)$,$(B)$,$(C)$,and $(D)$ are all correct. Given the options,$(C)$ is the best fit.

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

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