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:

1.	\(4\) ms-1	2.	\(-12\) ms−1
3.	\(42\) ms−1	4.	\(-9\) ms−1

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

A body is moving according to the equation $x=at+b{t}^2-c{t}^3$ where x = displacement and a, b and c are constants. The acceleration of the body is 

1. $a+2bt$ 

2. $2b+6ct$ 

3. $2b-6ct$ 

4. $3b-6c{t}^2$ 

A particle travels 10 m in first 5 sec and 10m in the next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec ?

1. 8.3 m

2. 9.3 m

3. 10.3 m

4. None of above