A body starts from rest and travels a distance $S$ with uniform acceleration, then moves uniformly a distance $2S$ and finally comes to rest after moving further $3S$ under uniform retardation. The ratio of the average velocity to maximum velocity is

A person travels $x$ distance with velocity $v_1$ and then $x$ distance with velocity $v_2$ in the same direction. The average velocity of the person is $v$, then the relation between $v , v _1$ and $v _2$ will be :

A particle moving in a straight line covers half the distance with speed $6 \mathrm{~m} / \mathrm{s}$. The other half is covered in two equal time intervals with speeds 9 $\mathrm{m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$ respectively. The average speed of the particle during the motion is :

The average velocity of a body moving with uniform acceleration travelling a distance of $3.06\, m$ is $ 0.34 ms^{-1}$. If the change in velocity of the body is $ 0.18ms^{-1}$ during this time, its uniform acceleration is.........$ms^{-2}$

A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the partk. Then cycles along the circumference and returns to the centre along $QO$ as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour respectively) is

The speed-time graph of a particle moving along a fixed direction is shown in Figure 

$(a)\; t = 0\; s$ to $10 \;s$, $(b)\;t=2 \;s$ to $6\; s$

What is the average speed of the particle over the intervals in $(a)$ and $(b)$?