A stone falls freely under gravity. It covers distances \(h_1,~h_2\) and \(h_3\) in the first \(5\) seconds, the next \(5\) seconds and the next \(5\) seconds respectively. The relation between \(h_1,~h_2\) and \(h_3\) is:

1.	\(h_1=\frac{h_2}{3}=\frac{h_3}{5}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)
2.	\(h_2=3h_1\) and \(h_3=3h_2\)
3.	\(h_1=h_2=h_3\)
4.	\(h_1=2h_2=3h_3\)

A particle moves a distance \(x\) in time \(t\) according to equation \(x=(t+5)^{-1}.\) The acceleration of the particle is proportional to:
1. (velocity)^\(3/2\)
2. (distance)^\(2\)
3. (distance)^\(-2\)
4. (velocity)^\(2/3\)

A particle starts its motion from rest under the action of a constant force. If the distance covered in the first \(10\) s is \(S_1\) and that covered in the first \(20\) s is \(S_2\), then:
1. \(S_2=2S_1\)
2. \(S_2 = 3S_1\)
3. \(S_2 = 4S_1\)
4. \(S_2= S_1\)

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

         

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

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

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. \(\frac{1}{2}f_0T\)