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 $\mathrm{S}={\mathrm{t}}^3-6{\mathrm{t}}^2+3\mathrm{t}+4$ 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:
1. \(\frac{a}{b}\)
2. \(\frac{2a}{3b}\)
3. \(\frac{a}{3b}\)
4. zero

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) $3b-6c{t}^2$