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:

1.	\(50\) m	2.	\(48\) m
3.	\(100\) m	4.	\(150\) m

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:

The figure gives the \((x\text-t)\) plot of a particle in a one-dimensional motion. Three different equal intervals of time are shown. The signs of average velocity for each of the intervals \(1,\) \(2\) & \(3,\) respectively are:

A body in one-dimensional motion has zero speed at an instant. At that instant, it must have:

When the velocity of a body is variable, then:

A body is projected vertically in the upward direction from the surface of the earth. If the upward direction is taken as positive, then the acceleration of the body during its upward and downward journey is:

Which one of the following displacement-time graph represents two moving objects \(P\) and \(Q\) with zero relative velocity?

1.		2.
3.		4.

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 \(g = 10~\text{m/s}^2)\)$$

A body is thrown upwards and reaches its maximum height. At that position:

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 \(B\):