Suppose you are riding a bike with a speed of \(20\) m/s due east relative to a person \(A\) who is walking on the ground towards the east. If your friend \(B\) walking on the ground due west measures your speed as \(30\) m/s due east, find the relative velocity between two reference frames \(A\) and \(B\):

1.	The velocity of \(A\) with respect to \(B\) is \(5\) m/s towards the east.
2.	The velocity of \(A\) with respect to \(B\) is \(5\) m/s towards the west.
3.	The velocity of \(A\) with respect to \(B\) is \(10\) m/s towards the east.
4.	The velocity of \(A\) with respect to \(B\) is \(10\) m/s towards the west.

A particle is moving along the \(x\)-axis such that its velocity varies with time as per the equation \(v = 20\left(1-\frac{t}{2}\right)\). At \(t=0\) particle is at the origin. From the following, select the correct position \((x)\) - time \((t)\) plot for the particle:

1.		2.
3.		4.

The velocity \(v\) of an object varies with its position \(x\) on a straight line as \(v=3\sqrt{x}.\) Its acceleration versus time \((a\text-t)\) graph is best represented by:

1.		2.
3.		4.

The displacement \((x)\) of a point moving in a straight line is given by; \(x=8t^2-4t.\) Then the velocity of the particle is zero at:

An elevator whose floor to ceiling height is \(12\) meters, moves upward with an acceleration of \(2.2~\text{m/s}^2\). After \(1.5\) seconds since starting, a bolt falls from its ceiling. The time taken by the bolt to reach the floor is:
1. \(1~\text{s}\)
2. \(2~\text{s}\)
3. \(\sqrt{2}~\text{s}\)
4. \(\sqrt{3}~\text{s}\)

A ball is thrown vertically downwards with a velocity of \(20\) m/s from the top of a tower. It hits the ground after some time with the velocity of \(80\) m/s . The height of the tower is: (assuming \(g = 10~\text{m/s}^2)\)$$

The position (\(x\)) of a particle in a straight line motion is given by \(x = 2 + 10 t - 5 t^{2}~\text{m}\). Its velocity (\(v\)) is best represented by?

1.		2.
3.		4.

A body starting from rest moves with uniform acceleration on a horizontal surface. The body covers \(3\) consecutive equal distances from the beginning in time \(t_1, t_2,\text{and}~t_3\)${\mathrm{}}_{}$ seconds. The ratio of \(t_1:t_2:t_3\) is:
1. \(1:2:3\)
2. \(1:\sqrt{2}:\sqrt{3}\)
3. \(1:\left(\sqrt{2}-1\right):\left(\sqrt{3}-\sqrt{2}\right)\)
4. \(\sqrt{3}:\sqrt{2}:1\)$\mathrm{}$

A particle starts from rest (with constant acceleration) and acquires velocity \(20\) m/s in \(5\) s. The distance travelled by the particle in the next \(2\) s will be:

A drunkard walking in a narrow lane takes \(5\) steps forward and \(3\) steps backward, followed again by \(5\) steps forward and \(3\) steps backward, and so on. Each step is \(1\) m long and requires \(1\) s. There is a pit on the road \(13\) m away from the starting point. The drunkard will fall into the pit after: