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) ${\mathrm{s}}_2={\mathrm{s}}_1$

The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3}m{s}^{-2},$ in the third second is 

(1) 6m                             

(2) 4m

(3) $\frac{10}{3}m$                         

(4) $\frac{19}{3}m$

A particle shows a distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point:

1. B

2. C

3. D

4. A

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 $m{s}^{-1}$to $20 m{s}^{-1}$ while passing through a distance of 135 m in t seconds. The value of t is:

(1) 10

(2) 1.8 

(3) 12

(4) 9

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 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

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: