If a body travels some distance in a given time interval, then for that time interval, its:

1.	Average speed  ≥  |Average velocity|
2.	|Average velocity|  ≥  Average speed
3.	Average speed < |Average velocity|
4.	|Average velocity| must be equal to average speed.

A body starts falling from height \(h\) and if it travels a distance of \(\frac{h}{2}\) during the last second of motion, then the time of flight is (in seconds):
1. \(\sqrt{2}-1\)
2. \(2+\sqrt{2}\)
3. \(\sqrt{2}+\sqrt{3}\)
4. \(\sqrt{3}+2\)

For a particle, displacement time relation is given by $t$ $=$ $\sqrt{x}$ $+$ $3$ . Its displacement, when its velocity is zero will be:
1. \(2\) m
2. \(4\) m
3. \(0\)
4. none of the above

A particle starts from rest with constant acceleration. The ratio of space-average velocity to the time-average velocity is:
where time-average velocity and space-average velocity, respectively, are defined as follows: 

$<v{>}_{time}$ $=$ $\frac{\int vdt}{\int dt}$
$<v{>}_{space}$ $=$ $\frac{\int vds}{\int ds}$

If a ball is thrown vertically upwards with speed \(u\), the distance covered during the last \(t\) seconds of its ascent is:
1. \(ut\)
2. \(\frac{1}{2}gt^2\)
3. \(ut-\frac{1}{2}gt^2\)
4. \((u+gt)t\)

A man throws some balls with the same speed vertically upwards one after the other at an interval of \(2\) seconds. What should be the speed of the throw so that more than two balls are in the sky at any time? (Given \(g = 9.8\) m/s²)

A car is moving along a straight line, say OP in the figure. It moves from O to P in \(18\) s and returns from P to Q in \(6.0\) s. The average velocity and average speed of the car in going from O to P and back to Q respectively are:

1. \(10\) m/s & \(10\) m/s

2. \(20\) m/s & \(30\) m/s 

3. \(20\) m/s & \(20\) m/s

4. \(10\) m/s & \(20\) m/s

The position of an object moving along \(x\)-axis is given by \(x=a+bt^2\), where \(a=8.5\) m, \(b=2.5\) ${\mathrm{ms}}^{-2}$ and \(t\) is measured in seconds. Its average velocity between \(t=2.0\) s and \(t=4.0\) s is:
1. \(10\) m/s
2. \(15\) m/s
3. \(20\) m/s
4. \(25\) m/s

A ball is thrown vertically upwards with a velocity of \(20\) m/s from the top of a multistorey building. The height of the point from where the ball is thrown is \(25.0\) m from the ground. How long will it be before the ball hits the ground? (Take \(g=10\) ms^–2.)
1. \(3\) s
2. \(2\) s
3. \(5\) s
4. \(20\) s

The position of an object moving along the x-axis is given by, \(x=a+bt^2\)  where \(a=8.5\) m,  \(b=2.5\) ms^–2, 
and \(t\) is measured in seconds. Its velocity at \(t=2.0\) s will be:
1. \(13\) m/s
2. \(17\) m/s
3. \(10\) m/s
4. \(0\)