For a particle in uniform circular motion, the acceleration $\overrightarrow{ a }$ at any point $P ( R , \theta)$ on the circular path of radius $R$ is (when $\theta$ is measured from the positive $x\,-$axis and $v$ is uniform speed)
$-\frac{v^{2}}{R} \sin \theta \hat{i}+\frac{v^{2}}{R} \cos \theta \hat{j}$
$-\frac{v^{2}}{R} \cos \theta \hat{i}+\frac{v^{2}}{R} \sin \theta \hat{j}$
$-\frac{v^{2}}{R} \cos \theta \hat{i}-\frac{v^{2}}{R} \sin \theta \hat{j}$
$-\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}$
If the instantaneous velocity of a particle projected as shown in figure is given by $v =a \hat{ i }+(b-c t) \hat{ j }$, where $a, b$, and $c$ are positive constants, the range on the horizontal plane will be
A ball is projected with velocity $V_0$ at an angle of elevation $30^o $ . Mark the correct statement
A particle has initial velocity $(3\hat i + 4\hat j$$ ) $ and has acceleration $(0.4\,\hat i + 0.3\,\hat j)$ . Its speed after $10\,s$ is
A particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)
A particle is moving on a circular path of radius $r$ with uniform velocity $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = {40^o})$