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

$1.$ $3\mathrm{A}+7\mathrm{B}$
$2.$ $\frac{3\mathrm{A}}{2}+\frac{7\mathrm{B}}{3}$
$3.$ $\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{3}$
$4.$ $\frac{\mathrm{A}}{3}+\frac{\mathrm{B}}{2}$

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by $x_p\left(t\right)=at+b{t}^2$ and $x_Q\left(t\right)=ft-t^2$. At what time do the cars have the same velocity?

1. $\frac{a-f}{1+b}$

2. $\frac{a+f}{2\left(b-1\right)}$

3. $\frac{a+f}{2\left(1+b\right)}$

4. $\frac{f-a}{2\left(1+b\right)}$

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time \(t_1.\) On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time \(t_2.\) The time taken by her to walk upon the moving escalator will be:

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 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