The figure shows a velocity-time graph of a particle moving along a straight line  If the particle starts from the position $x_0=-15\,m$ , then its position at $t=2\,s$ will be   ........ $m$

A particle moves from the point $\left( {2.0\hat i + 4.0\hat j} \right)\,m$, at $t = 0$ with an initial velocity $\left( {5.0\hat i + 4.0\hat j} \right)\,m{s^{ - 1  }}$. It is acted upon by a constant force which produces a constant acceleration $\left( {4.0\hat i + 4.0\hat j} \right)\,m{s^{ - 2}}$. What is the distance of the particle from the origin at time $2\,s$

Let $\vec v$ and $\vec a$ denote the velocity and acceleration respectively of a body in one-dimensional motion 

A particle moves along an arc of a circle of radius $R$ . Its velocity depends on the distance covered as $v = a\sqrt s$ , where $a$ is a constant then the angle $\alpha $ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be

A particle moves such that its position vector $\overrightarrow{\mathrm{r}}(\mathrm{t})=\cos \omega \mathrm{t} \hat{\mathrm{i}}+\sin \omega \mathrm{t} \hat{\mathrm{j}}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\overrightarrow{\mathrm{v}}(\mathrm{t})$ and acceleration $\overrightarrow{\mathrm{a}}(\mathrm{t})$ of the particle