The displacement of a particle is given by \(y = a + bt + ct^{2} - dt^{4}\). The initial velocity and acceleration are, respectively:

1.	\(b, -4d\)	2.	\(-b,2c\)
3.	\(b, ~2c\)	4.	\(2c, -2d\)

The position \(x\) of a particle varies with time \(t\) as \(x=at^2-bt^3\). The acceleration of the particle will be zero at time \(t\) equal to:

A student is standing at a distance of \(50\) metres from the bus. As soon as the bus begins its motion with an acceleration of \(1\) ms^–2, the student starts running towards the bus with a uniform velocity \(u\). Assuming the motion to be along a straight road, the minimum value of \(u\), so that the student is able to catch the bus is:
1. \(5\) ms^–1
2. \(8\) ms^–1
3. \(10\) ms^–1
4. \(12\) ms^–1

A body starts to fall freely under gravity. The distances covered by it in the first, second and third second will be in the ratio: 

A particle moving in a straight line covers half the distance with a speed of \(3~\text{m/s}\). The other half of the distance is covered in two equal time intervals with speeds of \(4.5~\text{m/s}\) and \(7.5~\text{m/s}\) respectively. The average speed of the particle during this motion is:
1. \(4.0~\text{m/s}\)
2. \(5.0~\text{m/s}\)
3. \(5.5~\text{m/s}\)
4. \(4.8~\text{m/s}\)

A stone dropped from a building of height \(h\) and reaches the earth after \(t\) seconds. From the same building, if two stones are thrown (one upwards and other downwards) with the same velocity \(u\) and they reach the earth surface after \(t_1\) and \(t_2\) seconds respectively, then: 

1. $t=t_1-t_2$

2. $t=\frac{t_1+t_2}{2}$

3. $t=\sqrt{t_1{t}_2}$

4. $t=t_1^2{t}_2^2$ 

The velocity-time \((v\text-t)\) graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in \(6\) s are, respectively: 

1. \(8\) m, \(16\) m
2. \(16\) m, \(8\) m
3. \(16\) m, \(16\) m
4. \(8\) m, \(8\) m

In the following graph, the distance travelled by the body in metres is:

Given below are two statements: