The velocity of a body at time t = 0 is 10sqr | vedclass

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(A) Given: Initial velocity $\vec{u} = 10\sqrt{2} \, m/s$ (North-East),Acceleration $\vec{a} = 2 \, m/s^2$ (South),Time $t = 5 \, s$.Let the East dire

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$10 \, m/s$,towards east

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(A) Given: Initial velocity $\vec{u} = 10\sqrt{2} \, m/s$ (North-East),Acceleration $\vec{a} = 2 \, m/s^2$ (South),Time $t = 5 \, s$.Let the East direction be the $x$-axis and the North direction be the $y$-axis.Initial velocity $\vec{u} = 10\sqrt{2} \cos(45^\circ) \hat{i} + 10\sqrt{2} \sin(45^\circ) \hat{j} = 10 \hat{i} + 10 \hat{j} \, m/s$.Acceleration $\vec{a} = 0 \hat{i} - 2 \hat{j} \, m/s^2$.Final velocity $\vec{v} = \vec{u} + \vec{a}t = (10 \hat{i} + 10 \hat{j}) + (0 \hat{i} - 2 \hat{j}) 	imes 5 = 10 \hat{i} + (10 - 10) \hat{j} = 10 \hat{i} \, m/s$.The magnitude of velocity is $v = |10 \hat{i}| = 10 \, m/s$.The direction is along the positive $x$-axis,which is towards the East.

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

The velocity of a body at time $t = 0$ is $10\sqrt{2} \, m/s$ in the north-east direction and it is moving with an acceleration of $2 \, m/s^2$ directed towards the south. The magnitude and direction of the velocity of the body after $5 \, s$ will be

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