BM Sharma Solutions for Chapter: Motion in One Dimension, Exercise 2: DPP

The relation between time and distance is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. The retardation is

The relation $3t=\sqrt{3x}+6$ describes the displacement of a particle in one direction, where $x$ is in $\mathrm{meter}$ and $t$ in $\mathrm{second}$. The displacement when the velocity is zero is

If the velocity of a particle is given by $v={(180-16x)}^{1/2 } {\text{m s}}^{-1}$, then its acceleration will be

A particle moves along $x-$axis as $x=4\left(t-2\right)+a{\left(t-2\right)}^2$. Which of the following is true? Here $a$ is a positive constant.

The position vector of a particle is given as $\overrightarrow{\mathit{r}}=\left(t^2-4t+6\right) \hat{\mathrm{i}}+\left(t^2\right) \hat{\mathrm{j}}$. The time after which the velocity vector and the acceleration vector become perpendicular to each other, is equal to

A particle of mass $m$ moves along a curve, $y=x^2$. When the particle has $x-$coordinate as $\frac{1}{2} \mathrm{m}$ and $x-$ component of velocity as$4 \mathrm{m} {\mathrm{s}}^{-1}$, then

The velocity of a particle moving in the positive direction of $x-$axis varies as $v=10\sqrt{x}$. Assuming that at $t=0$, particle was at $x=0$,

A particle moves along $x-$axis such that its position varies with time as $x=50t-5t^2$. Select the correct alternative(s).