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

The motion of a particle is described by the equation $x=a+b{t}^2$ where a = 15 cm and b = 3 cm/s². Its instantaneous velocity at time 3 sec will be 

1. 36 cm/sec

2. 18 cm/sec

3. 16 cm/sec

4. 32 cm/sec