$A$ particle starts from rest,accelerates at $2 \, m/s^2$ for $10 \, s$,then moves at a constant speed for $30 \, s$,and finally decelerates at $4 \, m/s^2$ until it stops. What is the total distance travelled by it in $m$?

At time $t=0$,a particle leaves the origin and moves in the positive direction of the $X$-axis. If the velocity of the particle varies as $v(t)=v_0(1-t/t_0)$,where $|v_0|=10 \ m/s$ and $t_0=10 \ s$,then the distance covered by the particle during the first $20 \ s$ is: (in $m$)

$A$ particle of mass $m$ moves on the $x$-axis as follows: it starts from rest at $t = 0$ from the point $x = 0$ and comes to rest at $t = 1$ at the point $x = 1$. No other information is available about its motion at intermediate time $(0 < t < 1)$. If $\alpha$ denotes the instantaneous acceleration of the particle,then

$A$ street car moves rectilinearly from station $A$ to the next station $B$ with an acceleration varying according to the law $a = (b - cx)$,where $b$ and $c$ are constants and $x$ is the distance from station $A$. The distance between the two stations and the maximum velocity are:

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

$A$ body of mass $0.40 \;kg$ moving initially with a constant speed of $10 \;m s^{-1}$ to the north is subject to a constant force of $8.0 \;N$ directed towards the south for $30 \;s$. Take the instant the force is applied to be $t=0$,the position of the body at that time to be $x=0$,and predict its position at $t=-5 \;s, 25 \;s, 100 \;s$.