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An object with a speed of 6.25 m/s, is decelerated at a rate given by:
$\frac{dv}{dt}$$=-2.5\sqrt{v}$ where v is the instantaneous speed. The time taken by object, to come to rest, would be:
1. 2 s
2. 4 s
3. 8 s
4. 1 s

A particle of mass m is at rest at the origin at time t=0. It is subjected to a force F(t) =$F_0{e}^{-bt}$ in the x-direction. Its speed v(t) is depicted by which of the following curves?
1. 
2. 
3. 
4. 

A point moves in a straight line so that its displacement x metre at time t sec is given by $x^2=1+t^2$. It is acceleration in $m/s^2$ at time t sec is 
1. $\frac{1}{x^3}$
2. $\frac{1}{x}-\frac{1}{x^2}$
3. $\frac{1}{x}-\frac{t^2}{x^3}$
4. $\frac{-t}{x^2}$

If position (in meter)  of a particle moving in straight line is given by $x=t^2-2t+1$ (where t is time in second). The distance travelled by particle in first two second is 
1. zero
2. 2 m
3. 4 m
4. 3 m

The displacement x of a particle varies with time t as $x=a{e}^{-\alpha t}+b{e}^{\beta t}$, where a, b, $\alpha and \beta$ are positive constants. The velocity of the particle is 
1. be independent of $\beta$
2. drop to zero when $\alpha =\beta$
3. go on decreasing with time
4. go on increasing with time

A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is
1. $4\sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$
2. $2\sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$
3. $\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$
4. $\sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$

Two particles start moving from the same point along the straight line. The first moves with constant velocity v and the second with constant acceleration a. During the time that elapses before the second catches the first, the greatest distance between the particles is 
1. $\frac{{\mathrm{v}}^2}{\mathrm{a}}$
2. $\frac{{\mathrm{v}}^2}{2\mathrm{a}}$
3.$\frac{2{\mathrm{v}}^2}{\mathrm{a}}$
4. $\frac{{\mathrm{v}}^2}{4\mathrm{a}}$

A particle is projected vertically upwards from a point A on the ground. It takes time t₁ to reach a point B , but it still continues to move up. If it takes further t₂ time to reach the ground from point B. Then height of point B from the ground is 
1. $\frac{1}{2}g{\left(t_1+t_2\right)}^2$
2. ${\mathrm{gt}}_1{\mathrm{t}}_2$
3. $\frac{1}{2}g{\left(t_1\mathit{-}{t}_2\right)}^2$
4. $\frac{1}{2}g{t}_1{t}_2$

A projectile of fired vertically upwards with an initial velocity u. After an interval of T seconds a second projectile is fired vertically upwards, also with initial velocity u. The correct statment is 
1. They meety at time $\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$
2. They meet at time $\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}+\frac{\mathrm{T}}{2}$
3. They meet at time $\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}-\frac{\mathrm{T}}{2}$
4. They never meet

A body starts from rest with uniform acceleration and remains in motion for n seconds. If its final velocity after n second is v, then its displacement in the last two seconds will be 
1. $\frac{2\mathrm{v}\left(\mathrm{n}+1\right)}{\mathrm{n}}$
2. $\frac{\mathrm{v}\left(\mathrm{n}+1\right)}{\mathrm{n}}$
3. $\frac{\mathrm{v}\left(\mathrm{n}-1\right)}{\mathrm{n}}$
4. $\frac{2\mathrm{v}\left(\mathrm{n}-1\right)}{\mathrm{n}}$

A body is in rectilinear motion with an acceleration given by $\mathrm{a}=2{\mathrm{v}}^{3/2}$. If particle starts its motion from origin with a velocity of 4 ms^-1 , the position of x of the particle at an instant in terms of v can be given as
1. $\frac{1}{\sqrt{\mathrm{v}}}=\frac{1}{2}-x$
2. $\sqrt{\mathrm{v}}=x+2$
3. $\sqrt{\mathrm{v}}=x$
4. $\sqrt{\mathrm{v}}=2x-1$

A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4 h. The two stones cross each other after some time 
1. $\sqrt{\frac{\mathrm{h}}{2\mathrm{g}}}$
2. $\sqrt{\frac{\mathrm{h}}{8\mathrm{g}}}$
3. $\sqrt{2\mathrm{hg}}$
4. $\sqrt{8\mathrm{hg}}$

The velocity of a particle that moves in the positive x-direction varies with its position (x) as shown in the figure. Its acceleration at x = 6 m is 

1. $-\frac{75}{4} \mathrm{m}/{\mathrm{s}}^2$
2. $-\frac{5}{4} \mathrm{m}/{\mathrm{s}}^2$
3. $-\frac{25}{2} \mathrm{m}/{\mathrm{s}}^2$
4. $-\frac{125}{8} \mathrm{m}/{\mathrm{s}}^2$

A train starts from station A with uniform acceleration a₁ for some distance and then goes with uniform retardation a₂ for some more distance to come to rest at station B. The distance between A and B is 4 km and the train takes 4 hours to complete this journey. If acceleration and retardation are in km/hour², then
1. $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=4$
2. $\frac{1}{{\mathrm{a}}_1}+\frac{1}{a_2}=2$
3. ${\mathrm{a}}_1{\mathrm{a}}_2=1$
4. None 

The velocity of a particle moving along the x-axis is given by $\mathrm{v}={\mathrm{x}}^3-6{\mathrm{x}}^2+12$ where v is in m/s and x is in m. Acceleration of the particle when it is passing through the point x = 4 m wil be 
1. -20 ${\mathrm{ms}}^{-2}$
2. -10 ${\mathrm{ms}}^{-2}$
3. -5 ${\mathrm{ms}}^{-2}$
4. zero 
 

For position-time (x-t) curve as shown in figure, the velocity-time (v-t) curve will be 

1. 
2. 
3. 
4. 

The graph given shows the velocity v versus time t for a body. Which of the following graphs shown represents the corresponding acceleration versus time graphs ?

1. 
2. 
3. 
4. 

A balloon is moving vertically upward with a velocity of 4 m/s. When its is at a height of h, a stone is dropped from it . If it reaches the ground in 4s, the height of the balloon, when the stone is released, is (g = 9.8 $\mathrm{m}/{\mathrm{s}}^2$)
1. 62.4 m
2. 42.4 m
3. 78.4 m
4. 82.2 m

What is the average velocity during time interval t =2 s to t = 5 s, in the following position time curve ?

1. 2 m/s
2. 2/3 m/s
3. 1.2 m/s
4. 0.4 m/s

Consider the given velocity-time graph. It represents the motion of 

1. a projectile projected vertically upward, from a point
2. an electron in the hydrogen atom
3. a bullet fired horizontally from the top of a tower
4. an object in the positive direction with decreasing speed