A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation 

(1) Both will be equal

(2) First will be half of second

(3) First will be 1/4 of second

(4) No definite ratio

A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second 

(1) $\frac{7}{5}$

(2) $\frac{5}{7}$

(3) $\frac{7}{3}$

(4) $\frac{3}{7}$ 

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 \(S = t^{3} - 6 t^{2} + 3 t + 4\) metres. The velocity when the acceleration is zero is:

If a body starts from rest and travels 120 cm in the 6^th second, then what is the acceleration 

(1) 0.20 m/s²

(2) 0.027 m/s²

(3) 0.218 m/s²

(4) 0.03 m/s²

If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s. Then it covers a distance of 

(1) 20 m

(2) 400 m

(3) 1440 m

(4) 2880 m

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:

If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

(1) 20 ms^–2

(2) 10 ms^–2

(3) 2 ms^–2

(4) 1 ms^–2

The displacement of a particle starting from rest (at t = 0) is given by $s=6t^2-t^3$. The time in seconds at which the particle will attain zero velocity again, is  

(1) 2

(2) 4

(3) 6

(4) 8

Two cars A and B are at rest at the same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with a constant acceleration of 4 m/s², then B will catch A after how much time?

(1) 10 sec

(2) 20 sec

(3) 30 sec

(4) 35 sec