A particle is moving such that its position coordinates (x, y) are (\(2\) m, \(3\) m) at time \(t=0,\) (\(6\) m,\(7\) m) at time \(t=2\) s, and (\(13\) m, \(14\) m) at time \(t=\) \(5\) s. The average velocity vector \(\vec{v}_{avg}\) from \(t=\) 0 to \(t=\) \(5\) s is:
1. \({1 \over 5} (13 \hat{i} + 14 \hat{j})\)
2. \({7 \over 3} (\hat{i} + \hat{j})\)
3. \(2 (\hat{i} + \hat{j})\)
4. \({11 \over 5} (\hat{i} + \hat{j})\)

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:

A particle has initial velocity \(\left(2 \hat{i} + 3 \hat{j}\right)\) and acceleration \(\left(0 . 3 \hat{i} + 0 . 2 \hat{j}\right)\). The magnitude of velocity after \(10\) s will be:

The motion of a particle along a straight line is described by the equation \(x = 8+12t-t^3\) where \(x \) is in meter and \(t\) in seconds. The retardation of the particle, when its velocity becomes zero, is:
1. \(24\) ms^-2
2. zero
3. \(6\) ms^-2
4. \(12\) ms^-2

A ball is dropped from a high-rise platform at \(t=0\) starting from rest. After \(6\) seconds, another ball is thrown downwards from the same platform with speed \(v\). The two balls meet after \(18\) seconds. What is the value of \(v\)?

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. \(S_2= S_1\)

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