The displacement x of a particle moving in one dimension under the action of a constant force is related to time t by the equation $t$ $=$ $\sqrt{x}$ $+$ $3$, where x is in metres and t is in seconds. What is the displacement of the particle from t = 0 s to t = 6 s?

1.  0

2.  12 m

3.  6 m

4.  18 m

The acceleration \(a\) (in ${\mathrm{ms}}^{-2}$) of a body, starting from rest varies with time \(t\) (in \(\mathrm{s}\)) as per the equation \(a=3t+4.\) The velocity of the body at time \(t=2\) \(\mathrm{s}\) will be:
1. \(10\) ${\mathrm{ms}}^{-1}$
2. \(18\) ${\mathrm{ms}}^{-1}$
3. \(14\) ${\mathrm{ms}}^{-1}$
4. \(26\) ${\mathrm{ms}}^{-1}$

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

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:

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

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