The displacement of a particle moving in a straight line depends on time as $x = \alpha t^3 + \beta t^2 + \gamma t + \delta$. The ratio of initial acceleration to its initial velocity depends on:

$A$ cyclist traversed half the distance of a linear track with a velocity $10 \ m \ s^{-1}$. The remaining part of the track was covered with a velocity $v_1$ for half the time and a velocity $v_2$ for the other half of the time. If $v_1+v_2=20 \ m \ s^{-1}$,then the average velocity of the cyclist during the completion of the journey through the track is (in $m \ s^{-1}$)

$A$ bogie of a uniformly moving train is suddenly detached from the train and stops after covering some distance. What is the relation between the distance covered by the bogie and the distance covered by the train in the same time?

Match the following columns.

Column $I$	Column $II$
$(A)$ Constant positive acceleration	$(p)$ Speed may increase
$(B)$ Constant negative acceleration	$(q)$ Speed may decrease
$(C)$ Constant displacement	$(r)$ Speed is zero
$(D)$ Constant slope of $a-t$ graph	$(s)$ Speed must increase

$A$ body is projected vertically up with a velocity $v$ and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight are

$A$ motorbike starts from rest, attains a velocity of $10 \,m/s$ with an acceleration of $0.5 \,m/s^2$, travels $10 \,km$ with this uniform velocity, and then comes to a halt with a uniform deceleration of $0.2 \,m/s^2$. The total time of travel is: (in $\,s$)