A scooter going due east at $10\, ms^{-1}$ turns right through an angle of $90^°$. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is

The figure shows the velocity and the acceleration of a point like body at the initial moment of its motion. The direction and the absolute value of the acceleration remain constant. Find the time when the speed becomes minimum.........$s$ (Given : $a = 4\, m/s^2, v_0 = 40\, m/s, \phi =143^o$)

Starting from the origin at time $t=0,$ with initial velocity $5 \hat{ j }\, ms ^{-1},$ a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{ i }+4 \hat{ j })\, ms ^{-2}$. At time $t$, its coordinates are $\left(20\, m , y _{0}\, m \right) .$ The values of $t$ and $y _{0},$ are respectively

Motion of a particle in $x - y$ plane is described by a set of following equations $x=4 \sin \left(\frac{\pi}{2}-\omega t\right) m$ and $y=4 \sin (\omega t) m$. The path of particle will be 

A particle starts from the origin at $\mathrm{t}=0$ with an initial velocity of $3.0 \hat{\mathrm{i}} \;\mathrm{m} / \mathrm{s}$ and moves in the $x-y$ plane with a constant acceleration $(6.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \;\mathrm{m} / \mathrm{s}^{2} .$ The $\mathrm{x}$ -coordinate of the particle at the instant when its $y-$coordinate is $32\;\mathrm{m}$ is $D$ meters. The value of $D$ is