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

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

An elevator car, whose floor to ceiling distance is equal to 2.7 m, starts ascending with constant acceleration of 1.2 ms^–2. 2 sec after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is 

1. $\sqrt{0.54}$ $s$

2. $\sqrt{6}$ $s$

3. 0.7 s

4. 1 s

The acceleration ‘a’ in m/s² of a particle is given by $a=3t^2+2t+2$ where t is the time. If the particle starts out with a velocity, u = 2 m/s at t = 0, then the velocity at the end of 2 seconds will be:

1. 12 m/s

2. 18 m/s

3. 27 m/s

4. 36 m/s

A particle moves along a straight line such that its displacement at any time t is given by $\mathrm{S}={\mathrm{t}}^3-6{\mathrm{t}}^2+3\mathrm{t}+4$ metres. The velocity when the acceleration is zero is:

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:
1. \(\frac{a}{b}\)
2. \(\frac{2a}{3b}\)
3. \(\frac{a}{3b}\)
4. zero

The relation \(3t = \sqrt{3x} + 6\) describes the displacement of a particle in one direction where \(x\) is in metres and \(t\) in seconds. The displacement, when velocity is zero, is: 

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

If the velocity of a particle is given by $v={(180-16x)}^{1/2}$ m/s, then its acceleration will be: 

1. Zero

2. 8 m/s²

3. – 8 m/s²

4. 4 m/s² 

Two trains, each 50 m long, are travelling in the opposite direction with velocities 10 m/s and 15 m/s. The time of crossing is:

1. 10 sec

2. 4 sec

3. $2\sqrt{3}$ $\sec$

4. $4\sqrt{3}$ $\sec$

The distance between two particles is decreasing at the rate of 6 m/sec when they are moving in the opposite directions. If these particles travel with the same initial speeds and in the same direction, then the separation increases at the rate of 4 m/sec. It can be concluded that particles' speeds could be

1. 5 m/sec, 1 m/sec

2. 4 m/sec, 1 m/sec

3. 4 m/sec, 2 m/sec

4. 5 m/sec, 2 m/sec