$A$ particle moves along the straight line $y=3x+5$. Which coordinate changes at a faster rate?

The equation of motion of a body is $\frac{dv}{dt} = -4v + 8$,where $v$ is the velocity in $m/s$ and $t$ is the time in seconds. The initial velocity of the particle was zero. Then,

$A$ 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 $t=-5 \;s, 25 \;s, 100 \;s$.

Two stones are thrown up simultaneously from the edge of a cliff $240 \ m$ high with initial speeds of $10 \ m/s$ and $40 \ m/s$ respectively. Which of the following graphs best represents the time variation of the relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance,take $g = 10 \ m/s^2$)

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 $m$)

$A$ parachutist drops freely from an aeroplane for $10\,s$ before the parachute opens. Then he descends with a net retardation of $2.5\,m/s^2$. If he bails out of the plane at a height of $2495\,m$ and $g = 10\,m/s^2$,his velocity on reaching the ground will be .......$m/s$.