An object starting from rest moves on a straight road for time $t$ and comes to rest finally. The distance is covered in two parts. In the first part, it is accelerated at constant acceleration $\alpha$ and then decelerate at rate $\beta$. Then maximum velocity is

Displacement-Time graph of the given Velocity-time graph is:

A body falling from rest describes distance ${\mathrm{S}}_1$, ${\mathrm{S}}_2$ and ${\mathrm{S}}_3$ in the $1^{\mathrm{st}}, 2^{\mathrm{nd}}$ and $3^{\mathrm{rd}}$ seconds of its fall. Then the ratio of ${\mathrm{S}}_1:{\mathrm{S}}_2:{\mathrm{S}}_3$ is:

A body starts from rest at time, $t=0$, the acceleration-time graph is shown in the figure. The maximum velocity attained by the body will be:  

A man walks along the footpath of a circular garden. During his motion along the circular path he makes several rounds. His net displacement will be zero whenever his distance travelled will be in multiples of:

The below given position-time graph describes which of the given options:

A teacher assigned a job of converting the position-time graph into a velocity-time graph. All the four Meera, Meena, Trisha and Akshaya plotted the graphs as under. The person to give the correct answer is:

A body starts from rest at $t=0$, the acceleration time graph is shown. The distance travelled by the body from $t=4 \mathrm{s}$ to $t=8 \mathrm{s}$ will be