The relation between time and distance is given by $t=\alpha x^2+\beta x$, where α and β are constants. The retardation, as calculated based on this equation, will be (assume v to be velocity) :

1. $2\alpha v^3$

2. $2\beta v^3$

3. $2\alpha \beta v^3$

4. $2{\beta }^2{v}^3$

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:

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. $4\sqrt{3}$ $\sec$