The relation between time $t$ and distance $x$ of a particle is $t=a x^2+b x$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

A car starts at time $t=0$ from an initial speed of $10 \mathrm{~m} / \mathrm{s}$ and accelerates uniformly with $2 \mathrm{~m} / \mathrm{s}^2$ on a straight road for time $0 \leq t \leq 10 \mathrm{~s}$. Let $s_1$ and $s_2$ be the distance covered by the car in time $3 \leq t \leq 4 \mathrm{~s}$ and $4 \leq t \leq 5 \mathrm{~s}$, respectively. The ratio $\frac{s_2}{s_1}$ is

A ball projected up passes the same height $H$ at 2 s and 10 s . The value of $H$ is (use, $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )

Two towns $X$ and $Y$ are connected by a regular bus service. A bus leaves in either direction at every $t=T$ minutes. A man moving with same speed in the direction $X$ to $Y$ find that a bus goes past him every $t=t_1$ minutes in the direction of his motion, and every $t=t_2$ minutes in the opposite direction. Then, $T$ is given by