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If v(t) = 3t-1 and x(2) = 1, then the original position function is: 
Hint: \(\left(v \left( t \right) = \frac{d s}{d t}\right)\)$$
1. $\frac{3}{2}{t}^2-t-3$
2. $\frac{1}{2}{t}^2-t-3$
3. $\frac{3}{2}{t}^2-2t-3$
4. None of the above

A car has a certain displacement between 0 seconds and 2 seconds. If we defined its velocity as v(t)=6t-5, then the displacement in meters is: $\left(\mathrm{Here}, \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$
1. 1 m
2. 2 m
3. 3 m
4. 4 m

If the distance 's' travelled by a body in time 't' is given by $\mathrm{s}=\frac{\mathrm{a}}{\mathrm{t}}+{\mathrm{bt}}^2$ then the acceleration equals
(1)  $\frac{2\mathrm{a}}{{\mathrm{t}}^3}+2\mathrm{b}$
(2)  $\frac{2\mathrm{s}}{{\mathrm{t}}^3}$
(3)  $2\mathrm{b}-\frac{2\mathrm{a}}{{\mathrm{t}}^3}$
(4)  $\frac{\mathrm{s}}{{\mathrm{t}}^2}$

A particle moves along a straight line such that at time t its position from a fixed point O on the line is $3{\mathrm{t}}^2-2$. The velocity of the particle when t=2 is:
(A)  8 ms^-1
(B)  4 ms^{-1
(C)  12 ms-1
(D)  0}

If the velocity of a particle moving on x-axis is given by $\mathrm{v}=3{\mathrm{t}}^2-12\mathrm{t}+6$. At which time is the acceleration of particle zero?
1.  $2$ sec
2.  $2+\sqrt{2}$ sec
3.  $2-\sqrt{2}$ sec
4.  zero

The acceleration of a particle is given by \(a=3t\) at \(t=0\), \(v=0\), \(x=0\). The velocity and displacement at \(t = 2~\text{sec}\) will be:
\(\left(\text{Here,} ~a=\frac{dv}{dt}~ \text{and}~v=\frac{dx}{dt}\right)\)
1. \(6~\text{m/s}, 4~\text{m}\)
2. \(4~\text{m/s}, 6~\text{m}\)
3. \(3~\text{m/s}, 2~\text{m}\)
4. \(2~\text{m/s}, 3~\text{m}\)

The velocity-time relation of an electron starting from rest is given by $\mathrm{v} =\mathrm{kt}$ where $\mathrm{k} =2 {\mathrm{ms}}^{–2}$. The distance traveled in 3 seconds is
1.  9 m
2.  27 m
3.  36 m
4.  16 m