A person travelling in a straight line moves with a constant velocity \(v_1\) for a certain distance \(x\) and with a constant velocity \(v_2\) ${\mathrm{}}_{}$for the next equal distance. The average velocity \(v\) is given by the relation:

1.	\(\frac{1}{v} = \frac{1}{v_1}+\frac{1}{v_2}\)	2.	\(\frac{2}{v} = \frac{1}{v_1}+\frac{1}{v_2}\)
3.	\(\frac{v}{2} = \frac{v_1+v_2}{2}\)	4.	\(v = \sqrt{v_1v_2}\)

A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time $t_1$ if the elevator is moving uniformly and time $t_2$ if the elevator is stationary. Then:

Two bodies, \(A\) (of mass \(1~\text{kg}\)) and \(B\) (of mass \(3~\text{kg}\)) are dropped from heights of \(16~\text{m}\) and \(25~\text{m}\), respectively. The ratio of the time taken by them to reach the ground is:
1. \(\frac{5}{4}\)
2. \(\frac{12}{5}\)
3. \(\frac{5}{12}\)
4. \(\frac{4}{5}\)

A particle moves along a straight line OX. At a time t (in seconds), the displacement x (in metres) of the particle from O is given by x= 40 + 12t – t³. How long would the particle travel before coming to rest?

A particle moving along the x-axis has acceleration \(f,\) at time \(t,\) given by, \(f=f_0\left ( 1-\frac{t}{T} \right ),\)  where \(f_0\) and \(T\) are constants. The particle at \(t=0\) has zero velocity. In the time interval between \(t=0\) and the instant when \(f=0,\) the particle’s velocity \( \left ( v_x \right )\) is:
1. \(f_0T\)
2. \(\frac{1}{2}f_0T^{2}\)
3. \(f_0T^2\)
4. \(\frac{1}{2}f_0T\)

A car moves from \(\mathrm{X}\) to \(\mathrm{Y}\) with a uniform speed \(\mathrm{v_u}\) and returns to \(\mathrm{X}\) with a uniform speed \(\mathrm{v_d}.\) The average speed for this round trip is:

1. $\frac{2{\mathrm{v}}_{\mathrm{d}}{\mathrm{v}}_{\mathrm{u}}}{{\mathrm{v}}_{\mathrm{d}}+{\mathrm{v}}_{\mathrm{u}}}$

2. $\sqrt{{\mathrm{v}}_{\mathrm{u}}{\mathrm{v}}_{\mathrm{d}}}$

3. $\frac{{\mathrm{v}}_{\mathrm{d}}{\mathrm{v}}_{\mathrm{u}}}{{\mathrm{v}}_{\mathrm{d}}+{\mathrm{v}}_{\mathrm{u}}}$

4. $\frac{{\mathrm{v}}_{\mathrm{u}}+{\mathrm{v}}_{\mathrm{d}}}{2}$

The position of a particle with respect to time \(t\) along the \(\mathrm{x}\)-axis is given by \(9t^{2}-t^{3}\) where x is in metre and \(t\) in second. What will be the position of this particle when it achieves maximum speed along the \(+\mathrm{x}\) direction?
1. \(32\) m
2. \(54\) m
3. \(81\) m
4. \(24\) m

A particle moves in a straight line with a constant acceleration. It changes its velocity from \(10\) ms^-1 to \(20\) ms^-1 while covering a distance of \(135\) m in \(t\) seconds. The value of \(t\) is:

A particle shows the distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point:

         

The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3}\) ms^-2, in the third second is:
1. \(6\) m
2. \(4\) m
3. \(\frac{10}{3}\) m
4. \(\frac{19}{3}\) m