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-axis is given by, \(x=a+bt^2\)  where \(a=8.5\) m, \(b=2.5\) ms^-2, and \(t\) is measured in seconds. Its velocity at \(t=2.0\) s will be:
1. \(13\) m/s
2. \(17\) m/s
3. \(10\) m/s
4. \(0\) 

The position x of a particle moving along the x-axis varies with time t as x = $20\mathrm{t}$ $-$ $5{\mathrm{t}}^2$, where x is in meters and t is in seconds. The particle reverses its direction of motion at:

1. x = 40 m

2. x = 10 m

3. x = 20 m

4. x = 30 m

A body in one-dimensional motion has zero speed at an instant. At that instant, it must have:

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 α and β are constants. The retardation, as calculated based on this equation, will be (assume v to be velocity) :

1. $2\alpha v^3$

2. $2\beta v^3$

3. $2\alpha \beta v^3$

4. $2{\beta }^2{v}^3$

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