Embibe Experts Solutions for Exercise 4: EXERCISE

The position of a particle moving in a straight line is given by $y=3t^3+t^2+5$ (where $y$ is in $\mathrm{cm}$ and $t$ is in second), then find the acceleration of the particle at time $t=2 \mathrm{s}.$

A particle moves in a straight line so that it covers a distance $s=\left(a{t}^3+bt+5\right)\mathrm{m},$ where $t$ is in second. If its acceleration after $4 \mathrm{s}$ is $48 \mathrm{m} {\mathrm{s}}^{–2},$ then $a$ is equal to

A particle moves in a straight line so that $s=\sqrt{t}$, then its acceleration is proportional to

A particle is located at $x=0$ and time $t=0$  , starts moving along the positive $x$ direction with a velocity $v$ that varies as $v=\alpha \sqrt{x}$ . The displacement of the particle varies with time as

A particle moves in a straight line in such a way that its velocity at any point is given by $v^2=2-3x$, where $x$ is measured from a fixed point. The acceleration is