The acceleration of a particle starting from rest varies with time according to the relation \(A= - a\omega^2\sin \omega t\). The displacement of this particle at a time \(t\) will be:

1.	\(-\frac{1}{2}\left(a\omega^2\sin\omega t\right)t^2\)	2.	\(a\omega \sin \omega t\)
3.	\(a\omega \cos \omega t\)	4.	\(a\sin \omega t\)

If the velocity of a particle is (10 + 2t²) m/s, then the average acceleration of the particle between 2 sec and 5 sec is:

(1) 2 m/s²

(2) 4 m/s²

(3) 12 m/s²

(4) 14 m/s²

A thief is running away on a straight road in a jeep moving with a speed of \(9\) m/s. A policeman chases him on a motorcycle moving at a speed of \(10\) m/s. If the instantaneous separation of the jeep from the motorcycle is \(100\) m, how long will it take for the policeman to catch the thief?
1. \(1\) s
2. \(19\) s
3. \(90\) s
4. \(100\) s

A car A is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h². After how much time will B catch up with A

(1) 1 hr

(2) 1/2 hr

(3) 1/4 hr

(4) 1/8 hr

The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become

(1) $\sqrt{3}u$

(2) $\sqrt{5}u$

(3) $\sqrt{11}u$

(4) $\sqrt{7}u$

Two trains one of length 100 m and another of length 125 m, are moving in mutually opposite directions along parallel lines, meet each other, each with speed 10 m/s. If their acceleration are 0.3 m/s² and 0.2 m/s² respectively, then the time they take to pass each other will be

1.  5 s
2.  10 s
3.  15 s
4.  20 s

A body starts from rest with uniform acceleration. If its velocity after n second is v, then its displacement in the last two seconds is

(1) $\frac{2v(n+1)}{n}$

(2) $\frac{v(n+1)}{n}$

(3) $\frac{v(n-1)}{n}$

(4) $\frac{2v(n-1)}{n}$

A particle is moving in a straight line and passes through a point O with a velocity of 6 ms^–1. The particle moves with a constant retardation of 2 ms^–2 for 4 s and there after moves with constant velocity. How long after leaving O does the particle return to O

(1) 3s

(2) 8s

(3) Never

(4) 4s

A particle is projected with velocity $v_0$ along x-axis. The deceleration of the particle is proportional to the square of the distance from the origin i.e., $a=\alpha x^2.$ The distance at which the particle stops is :

1.  $\sqrt{\frac{3v_0}{2\alpha }}$
2.  ${\left(\frac{{\displaystyle 3v_o}}{{\displaystyle 2\alpha }}\right)}^{\frac{1}{3}}$
3.  $\sqrt{\frac{3v{}_0{}^2}{2\alpha }}$
4.  ${\left(\frac{{\displaystyle 3v{}_0{}^2}}{{\displaystyle 2\alpha }}\right)}^{\frac{1}{3}}$  

A body is projected vertically up with a velocity v and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight are

(1) $\overrightarrow{v}/2$ and v/2

(2) 0 and v/2

(3) 0 and 0

(4) $\overrightarrow{v}/2$ and 0