If the velocity of a particle is $\mathrm{v}=\mathrm{At}+{\mathrm{Bt}}^2$, where A and B are constants, then the distance travelled by it between 1s and 2s is?

$1.$ $3\mathrm{A}+7\mathrm{B}$
$2.$ $\frac{3\mathrm{A}}{2}+\frac{7\mathrm{B}}{3}$
$3.$ $\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{3}$
$4.$ $\frac{\mathrm{A}}{3}+\frac{\mathrm{B}}{2}$

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x)=$\beta x^{-2n}$ where, $\beta$ and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

1. $-2n{\beta }^2{x}^{-2n-1}$

2. $-2n{\beta }^2{x}^{-4n-1}$

3. $-2{\beta }^2{x}^{-2n+1}$

4. $2n{\beta }^2{e}^{-4n}$⁺¹

A stone falls under gravity. It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h_2, and h₃ is

1. h₁=2h₂=3h₃

2. h₁=h₂/3=h₃/5

3. h₂=3h₁ and h3=3h₂

4. h₁=h₂=h₃

The motion of a particle along a straight line is described by the equation; \(x=8+12 t-t^3,\)  where \(x\) is in metre and \(t\) is in second. The retardation of the particle when its velocity becomes zero is:

A boy standing at the top of a tower of \(20\) m height drops a stone. Assuming \(g=10\) m/s², the velocity with which it hits the ground will be:
1. \(20\) m/s                                         
2. \(40\) m/s
3. \(5\) m/s                                           
4. \(10\) m/s

A particle covers half of its total distance with speed ${\mathrm{v}}_1$ and the rest half distance with speed ${\mathrm{v}}_2.$ Its average speed during the complete journey is 

(1) $\frac{{\mathrm{v}}_1{\mathrm{v}}_2}{{\mathrm{v}}_1+{\mathrm{v}}_2}$                                         

(2) $\frac{2{\mathrm{v}}_1{\mathrm{v}}_2}{{\mathrm{v}}_1+{\mathrm{v}}_2}$

(3) $\frac{2{\mathrm{v}}_1^2{\mathrm{v}}_2^2}{{\mathrm{v}}_1^2+{\mathrm{v}}_2^2}$                                         

(4)  $\frac{{\mathrm{v}}_1+{\mathrm{v}}_2}{2}$

A ball is dropped from a high rise platform at t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s. What is the value of v? (take g=10 $m{s}^{-2}$)

1. $75 m{s}^{-1}$                                   2. $55 m{s}^{-1}$

3. $40 m{s}^{-1}$                                   4. $60 m{s}^{-1}$

A particle moves a distance x in time t according to equation $x={\left(t+5\right)}^{-1}.$The acceleration of the particle is proportional to, 

1. ${\left(velocity\right)}^{3/2}$                                         
2. ${\left(dis\tan ce\right)}^2$
3. ${\left(dis\tan ce\right)}^{-2}$                                         
4. ${\left(velocity\right)}^{2/3}$

A bus is moving with a speed of \(10~\text{ms}^{-1}\) on a straight road. A scooterist wishes to overtake the bus in \(100~\text{s}\)$\mathrm{}$. If the bus is at a distance of \(1~\text{km}\) from the scooterist, with what minimum speed should the scooterist chase the bus?
1. \(20~\text{ms}^{-1}\)
2. \(40~\text{ms}^{-1}\)
3. \(25~\text{ms}^{-1}\)
4. \(10~\text{ms}^{-1}\)

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is ${\mathrm{s}}_1$ and that covered in the first 20 s is ${\mathrm{s}}_2,$ then

1. ${\mathrm{s}}_2=2{\mathrm{s}}_1$                                   
2. ${\mathrm{s}}_2=3{\mathrm{s}}_1$
3. ${\mathrm{s}}_2=4{\mathrm{s}}_1$                                     
4. ${\mathrm{s}}_2={\mathrm{s}}_1$