The displacement is given by $x=2t^2+t+5$, the acceleration at $t=2s$ is 

1. $4m/s^2$

2. $8m/s^2$

3. $10m/s^2$

4. $15m/s^2$

Two trains travelling on the same track are approaching each other with equal speeds of 40 m/s. The drivers of the trains begin to decelerate simultaneously when they are just 2.0 km apart. Assuming the decelerations to be uniform and equal, the value of the deceleration to barely avoid collision should be 

1. 11.8 m/s²

2. 11.0 m/s²

3. 1.6 m/s²

4. 0.8 m/s²

A body moves from rest with a constant acceleration of 5 m/s². Its instantaneous speed (in m/s) at the end of 10 sec is  

1. 50

2. 5

3. 2

4. 0.5

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