A person walks in such a way that he covers equal distance in each step. The person takes 2 steps forwards towards east, then takes a right turn and
walks 4 steps towards south, then takes a right turn and walks 6 steps towards west and then takes a right turn and walks further. The direction of his final position after a total of 20 steps walk with respect to his initial position is
Particle $A$ (which was located at the origin at time $t=0$ ) is moving along the $X$-axis with a constant speed of $1 \mathrm{~m} / \mathrm{s}$. Location of particle $B$ which is moving along the $Y$-axis is given by $y=c t^2$, where $c=1 \mathrm{~m} / \mathrm{s}^2$. Find the speed of particle $A$ relative to particle $B$ at $t=1 \mathrm{~s}$
A particle is moving in $X Y$-plane as $\mathbf{x}=\left(4 t+t^2\right) \hat{\mathbf{i}}$, $\mathbf{y}=\left(2 t+\frac{t^2}{2}\right) \hat{\mathbf{j}}$, where $\mathbf{x}$ and $\mathbf{y}$ are displacements measured along $X$ and $Y$-axes respectively, in metres and $t$ in seconds, What is the velocity of the particle?
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