The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$, then at $t = 2 \ s$, its position is: (in $m$)

$A$ car,moving with a speed of $50 \,km/hr$,can be stopped by brakes after at least $6 \,m$. If the same car is moving at a speed of $100 \,km/hr$,the minimum stopping distance is..........$m$

$A$ particle starts from rest at $x=0 \, m$ with an acceleration of $1 \, m/s^2$. At $t = 5 \, s$,it receives an additional acceleration in the same direction as its motion. At $t = 10 \, s$,its speed and position are $v$ and $x$,respectively. Had the additional acceleration not been provided,its speed and position would have been $v_0$ and $x_0$,respectively. It is found that $x - x_0 = 12.5 \, m$. Then one can conclude that $v - v_0$ is .............. $m/s$.

$A$ particle starts with an initial speed $u$ and retardation $a$ to come to rest in time $T$. The time taken to cover the first half of the total path travelled is .......

$A$ driver applies the brakes on seeing the red traffic signal $400 \ m$ ahead. At the time of applying brakes,the vehicle was moving with $15 \ m/s$ and retarding at $0.3 \ m/s^2$. The distance of the vehicle from the traffic light one minute after the application of brakes is: (in $m$)

The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,then what is the retardation?