A particle covers half of its total distance with speed ν₁ and the rest half distance with speed ν₂.
Its average speed during the complete journey is:

1. $\frac{v_1+v_2}{2}$

2. $\frac{v_1{v}_2}{v_1+v_2}$

3. $\frac{2v_1{v}_2}{v_1+v_2}$

4. $\frac{v_1^2{v}_2^2}{v_1^2+v_2^2}$

The motion of a particle is given by the equation \(S = \left(3 t^{3} + 7 t^{2} + 14 t + 8 \right) \text{m} ,\) The value of the acceleration of the particle at \(t=1~\text{s}\) is:

The displacement \(x\) of a particle varies with time \(t\) as \(x = ae^{-\alpha t}+ be^{\beta t}\)$\mathrm{}$, where \(a,\) \(b,\) \(\alpha,\) and \(\beta\) are positive constants. The velocity of the particle will:

A car is moving with velocity v. It stops after applying breaks at a distance of 20 m. If the velocity of the car is doubled, then how much distance it will cover (travel) after applying breaks?
1.  40 m
2.  80 m
3.  160 m
4.  320 m

A body starts falling from height \(h\) and if it travels a distance of \(\frac{h}{2}\) during the last second of motion, then the time of flight is (in seconds):
1. \(\sqrt{2}-1\)
2. \(2+\sqrt{2}\)
3. \(\sqrt{2}+\sqrt{3}\)
4. \(\sqrt{3}+2\)

For a particle, displacement time relation is given by; $t$ $=$ $\sqrt{x}$ $+$ $3$ . Its displacement, when its velocity is zero will be:
1. \(2\) m
2. \(4\) m
3. \(0\) m
4. none of the above

A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards the earth, their respective kinetic energies will be in the ratio of:

1. 1: 4

2. 1: 2

3. 1: $\sqrt{2}$

4. $\sqrt{2}$ :1

A particle starts from rest with constant acceleration. The ratio of space-average velocity to the time-average velocity is:
where time-average velocity and space-average velocity, respectively, are defined as follows: 
\(<v>_{time}\) \(=\) \(\frac{\int v d t}{\int d t}\)
\(<v>_{space}\) \(=\) \(\frac{\int v d s}{\int d s}\)$\frac{}{}$

A particle is thrown vertically upward. Its velocity at half its height is \(10\) m/s. Then the maximum height attained by it is: (Assume, \(g=\) \(10\) m/s²) 
1. \(8\) m
2. \(20\) m
3. \(10\) m
4. \(16\) m

If a ball is thrown vertically upwards with speed \(u\), the distance covered during the last \(t\) seconds of its ascent is:
1. \(ut\)
2. \(\frac{1}{2}gt^2\)
3. \(ut-\frac{1}{2}gt^2\)
4. \((u+gt)t\)