A particle moves in east direction with $15 \,m/sec$. for $2\, sec$ then moves northward with $5\, m/sec$. for  $8 \,sec$. then average velocity of the particle is  

A person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?

The displacement $x$ of a particle depend on time $t$ as $x = \alpha {t^{^2}} - \beta {t^3}$ 

$Assertion$ : 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

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

The figure shows the velocity $(v)$ of a particle plotted against time $(t)$