For the following acceleration versus time graph the corresponding velocity versus displacement graph is: 

                     

1.		2.
3.		4.

A stone falls freely from rest from a height h and it travels a distance $\frac{9h}{25}$ in the last second. The value of h is:

1. 145 m

2. 100 m

3. 125 m

4. 200 ms

A point moves in a straight line under the retardation \(av^2\)${\mathrm{}}^{}$. If the initial velocity is \(u,\) the distance covered in \(t\) seconds is:
1. \((aut)\)
2. \(\frac{1}{a}\mathrm{ln}(aut)\)
3. \(\frac{1}{a}\mathrm{ln}(1+aut)\)
4. \(a~\mathrm{ln}(aut)\)

A bullet loses \(\dfrac{1}{20}\) of its velocity passing through a plank. The least number of planks required to stop the bullet is: (All planks offers same retardation)
1. \(10\)
2. \(11\)
3. \(12\)
4. \(23\)

The velocity-time graph of a body is shown in the figure. For the intervals AB and BC, the ratio of the distances travelled by the body is- 

1. 3:1

2. 1:3

3. $\sqrt{3}:2$

4. None of these

A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by $(4t^3-2t)$, where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin ?

1. 28 m/s²

2. 22 m/s²

3. 12 m/s²

4. 10 m/s²

The acceleration of a moving body can be found from: 

1. Area under the velocity-time graph

2. Area under the distance-time graph

3. Slope of the velocity-time graph

4. Slope of the distance-time graph

A particle starting from rest moving with constant acceleration travels a distance x in first 2 seconds and a distance y in next two seconds, then  

1. y = x

2. y = 2x

3. y = 3x

4. y = 4x

The velocity of a body depends on time according to the equation $v=20+0.1t^2$. The body is undergoing 

1. Uniform acceleration

2. Uniform retardation

3. Non-uniform acceleration

4. Zero acceleration