If a body starts from rest and travels \(120\ \text{cm}\) in the \(6^{th}\) second, then what is the acceleration:

1. \(0.20\ \text{m/s}^2\)
2. \(0.027\ \text{m/s}^2\)
3. \(0.218\ \text{m/s}^2\)
4. \(0.03\ \text{m/s}^2\)

If a car at rest accelerates uniformly to a speed of \(144\ \text{km/h}\) in \(20\ \text{s}\). Then it covers a distance of:

1. \(20\ \text{m}\)
2. \(400\ \text{m}\)
3. \(1440\ \text{m}\)
4. \(2880\ \text{m}\)

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 \(72\ \text{km/h}\) is to be brought to rest in a distance of \(200\) metres, then its retardation should be:

1. \(20\ \text{ms}^{–2}\)
2. \(10\ \text{ms}^{–2}\)
3. \(2\ \text{ms}^{–2}\)
4. \(1\ \text{ms}^{–2}\) 

The displacement of a particle starting from rest (at \(t = 0\)) is given by \(𝑠 = 6 𝑡^ 2 − 𝑡^ 3\). 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 a uniform velocity of \(40\ \text{m/s}\) and \(B\) starts in the same direction with a constant acceleration of \(4\ \text{m/s}^2\), then \(B\) will catch \(A\) after how much time?

1. \(10\ \text{s}\)
2. \(20\ \text{s}\)
3. \(30\ \text{s}\)
4. \(35\ \text{s}\)

The motion of a particle is described by the equation \(𝑥 = 𝑎 + 𝑏 𝑡^ 2\) where \(a = 15\ \text{cm}\) and \(b = 3\ \text{cm/s}^2\). Its instantaneous velocity at time \(3\) seconds will be:

1. \(36\ \text{cm/s}\)
2. \(18\ \text{cm/s}\)
3. \(16\ \text{cm/s}\)
4. \(32\ \text{cm/s}\)

A body travels for \(15\) seconds starting from rest with constant acceleration. If it travels distances \(S_1,\ S_2\) and \(S_3\) in the first five seconds, the second five seconds and the next five seconds, respectively, the relation between  \(S_1,\ S_2\) and \(S_3\) is:

1. \(𝑆_ 1 = 𝑆 _2 = 𝑆 _3\)
2. \(5𝑆_ 1 =3 𝑆 _2 = 𝑆 _3\)$\mathrm{}$
3. \(𝑆_ 1 = \frac 13 𝑆 _2 = \frac 15 𝑆 _3\)${}_{}$
4. \(𝑆_ 1 =\frac 15 𝑆 _2 = \frac 13 𝑆 _3\)${}_{}$

A body is moving according to the equation \(𝑥 = 𝑎 𝑡 + 𝑏 𝑡^ 2 − 𝑐 𝑡^ 3\) where \(x\) is the displacement, and \(a,\ b\) and \(c\) are constants. The acceleration of the body is:

1. \(𝑎 + 2 𝑏 𝑡\) 
2. \(2 𝑏 + 6 𝑐 𝑡  \)
3. \(2 𝑏 − 6 𝑐 𝑡  \)
4. \(3 𝑏 − 6 𝑐 𝑡^ 2  \)

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