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(B) Given:Initial velocity $\vec{v}_1 = 5\,\hat{i}\,m/s$Final velocity $\vec{v}_2 = 5\,\hat{j}\,m/s$Time interval $\Delta t = 10\,s$Change in velocity

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$\frac{1}{\sqrt{2}}\,m/s^2$ toward north-west

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(B) Given:Initial velocity $\vec{v}_1 = 5\,\hat{i}\,m/s$Final velocity $\vec{v}_2 = 5\,\hat{j}\,m/s$Time interval $\Delta t = 10\,s$Change in velocity $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = (5\,\hat{j} - 5\,\hat{i})\,m/s$Magnitude of change in velocity $|\Delta \vec{v}| = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\,m/s$Average acceleration $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\,m/s^2$The direction of $\Delta \vec{v}$ is given by the vector $(5\,\hat{j} - 5\,\hat{i})$,which points toward the north-west direction.Thus,the average acceleration is $\frac{1}{\sqrt{2}}\,m/s^2$ toward north-west.

Column-$I$ (Angle of projection)	Column-$II$
$A. \theta = 45^{\circ}$	$1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$	$2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$	$3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$	$4. \frac{R}{H} = 4$

$A$ particle is moving eastwards with a velocity of $5\,m/s$. In $10\,s$,the velocity changes to $5\,m/s$ northwards. The average acceleration in this time is

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