The displacement \((x)\) of a point moving in a straight line is given by; \(x=8t^2-4t.\) Then the velocity of the particle is zero at:

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:

The position of an object moving along the \(x\text-\)axis is given by, \(x=a+bt^2\), where \(a=8.5 ~\text m,\)  \(b=2.5~\text{m/s}^2,\) and \(t\) is measured in seconds. Its velocity at \(t=2.0~\text s\) will be:
1. \(13~\text{m/s}\) 
2. \(17~\text{m/s}\)
3. \(10~\text{m/s}\)
4. \(0~\text{m/s}\)

The position \(x\) of a particle moving along the \(x\)-axis varies with time \(t\) as \(x=20t-5t^2,\) where \(x\) is in meters and \(t\) is in seconds. The particle reverses its direction of motion at:
1. \(x=40~\text{m}\)
2. \(x=10~\text{m}\)
3. \(x=20~\text{m}\)
4. \(x=30~\text{m}\)

Which of the following four statements is false?

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:

The relation between time and distance is given by $$\(t=\alpha x^2+\beta x,\) where \(\alpha\) and \(\beta\) are constants. The retardation, as calculated based on this equation, will be (assume \(v\) to be velocity):
1. $\mathrm{}$\(2\alpha v^3\)
2. \(2\beta v^3\)$\mathrm{}{\mathrm{}}^{}$
3. \(2\alpha\beta v^3\)$\mathrm{}{\mathrm{}}^{}$
4. \(2\beta^2 v^3\)$\mathrm{}{\mathrm{}}^{}$

If in one-dimensional motion, instantaneous speed \(v\) satisfies \(0\leq v<v_0,\) then: