$A$ train starting from rest first accelerates uniformly up to a speed of $80 \ km/h$ for time $t$,then it moves with a constant speed for time $3t$. The average speed of the train for this duration of journey will be (in $km/h$):

$A$ particle moving along a straight line covers the first half of the distance with a speed of $3 \, m \, s^{-1}$. The other half of the distance is covered in two equal time intervals with speeds of $4.5 \, m \, s^{-1}$ and $7.5 \, m \, s^{-1}$ respectively. The average speed of the particle during the motion is:

Two trains '$A$' and '$B$' of length '$l$' and '$4l$' are travelling into a tunnel of length '$L$' on parallel tracks from opposite directions with velocities $108\,km/h$ and $72\,km/h$,respectively. If train '$A$' takes $35\,s$ less time than train '$B$' to cross the tunnel,then the length '$L$' of the tunnel is $...........\,m$. (Given $L = 60l$)

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 $m$)

When the velocity of a body is variable,then:

$A$ charged object is launched inside a time-varying electric field. Its motion is recorded by a video camera on a video tape. When it is at a certain moment $A$,its position vector $\vec{r}$,velocity $\vec{v}$,and acceleration $\vec{a}$ are measured. $A$ student watches the video at a later time but mistakenly plays the tape in the reverse direction. What are the position,velocity,and acceleration of the object at moment $A$ observed by the student,respectively?