The coordinate of an object is given as a function of time by \(x = 7 t - 3 t^{2}\), where \(x\) is in metres and \(t\) is in seconds. Its average velocity over the interval \(t=0\) to \(t=4\) is will be:
1. \(5\) m/s
2. \(-5\) m/s
3. \(11\) m/s
​​​​​​​4. \(-11\) m/s

A particle moves along a straight line and its position as a function of time is given by$\mathrm{}$ \(x= t^3-3t^2+3t+3\) then particle:

A particle moves in a straight line, according to the law $\mathrm{x}=4\mathrm{a}\left[\mathrm{t}+\mathrm{a}<mspace\; width="1em"></mspace>\sin \left(\frac{\mathrm{t}}{\mathrm{a}}\right)<mspace\; width="1em"></mspace>\right]$, where x is its position in meters, t is in sec & a is some constant, then the velocity is zero at :

(1)  $\mathrm{x}=4{\mathrm{a}}^2\mathrm{\pi }<mspace\; width="1em"></mspace>\mathrm{meters}$

(2)  $\mathrm{t}=\mathrm{\pi }<mspace\; width="1em"></mspace>\sec$

(3)  $\mathrm{t}=0<mspace\; width="1em"></mspace>\sec$

(4)  none

A point moves in a straight line so that its displacement is x m at time t sec, given by ${\mathrm{x}}^2={\mathrm{t}}^2+1$. Its acceleration in $\mathrm{m}/{\mathrm{s}}^2$ at time 1 sec is:

(1)  $\frac{1}{\mathrm{x}}$

(2)  $\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^3}$

(3)  $\frac{1}{{\mathrm{x}}^3}$

(4)  $-\frac{{\mathrm{t}}^2}{{\mathrm{x}}^3}$

If the velocity of a particle is \(v=At+Bt^{2},\) where \(A\) and \(B\) are constants, then the distance travelled by it between \(1~\text{s}\) and \(2~\text{s}\) is:

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 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)^\(2/3\)