A body is moving with velocity $30\; m/s$ towards east. After $10$ seconds its velocity becomes $40\; m/s$ towards north. The average acceleration of the body is ...... $m/s^2$

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

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 $2\,s$

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 

$Assertion$ : A tennis ball bounces higher on hills than in plains.
$Reason$ : Acceleration due to gravity on the hill is greater than that on the surface of earth