A particle moves in a straight line under an acceleration varying linearly with time. Its velocity-time graph is shown in figure. Then

A particle starts from rest at $x=0$ and acceleration varies with $x$ as shown in the figure. Then

Two stones are thrown vertically upwards simultaneously from the same point on the ground with velocities $u_1=40 \mathrm{m} {\mathrm{s}}^{-1}$ $u_2=30 \mathrm{m} {\mathrm{s}}^{-1}$. Which of the following curve(s) represent(s) correct variation with time $t$ ? (Assume that stone does not bounce back after collision with ground)

$\left(y_1-y_2\right)=$ relative separation between stones
$\left(v_1-v_2\right)=$ relative velocity of two stones
$\left(a_1-a_2\right)=$ relative acceleration of two stones

A particle starts moving along$\mathrm{x}$-axis from origin with initial velocity of $3 \mathrm{m} {\mathrm{s}}^{-1}$. The acceleration a of the particle varies with position $\mathrm{x}$ as $\mathrm{a}={\mathrm{Kx}}^2$, where $\mathrm{K}$ is a constant. At $\mathrm{x}=\sqrt{3} \mathrm{m}$, the tangent to the graph makes an angle of $60°$ with $\mathrm{x}$-axis. Then at $\mathrm{x}=\sqrt{3} \mathrm{m}$ ( $\mathrm{v}$ represents velocity of particle at position $\mathrm{x}$ )

For the $v-t$ graph shown in figure match the following columns.