$A$ particle has an initial velocity $(2\hat i + 3\hat j)$ and an acceleration $(0.3\hat i + 0.2\hat j)$. Its speed after $10\,s$ is:

$A$ point moves in the $xy$-plane according to the following equations: $x = a \sin \omega t$ and $y = a(1 - \cos \omega t)$, where $a$ and $\omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.

$A$ particle is moving with velocity $\vec v = K(y\hat i + x\hat j)$ where $K$ is a constant. The general equation for its path is

If two bodies $A$ and $B$ are projected with the same velocity $u$ but with different angles $\theta_1$ and $\theta_2$ respectively with the horizontal such that both have the same range,then the ratio of the times of flight of the bodies $A$ and $B$ is:

$A$ particle moves from the point $(2.0\hat i + 4.0\hat j) \, m$ at $t = 0$ with an initial velocity $(5.0\hat i + 4.0\hat j) \, m/s$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat i + 4.0\hat j) \, m/s^2$. What is the distance of the particle from the origin at time $t = 2 \, s$?

When a body moves with a constant speed along a circle,