The distance travelled by a particle is proportional to the squares of time, then the particle travels with

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p-1)$ seconds and $S_2$ in first $p$ seconds. The displacement $S_1+S_2$ will be made in time :

Find the acceleration of particle at $x = 5\,m$ with the help of graph. where $v-$ velocity and $x-$ displacement

Position $x$ of a particle at any instant is related with velocity as $v = \sqrt {2x + 9}$ . The particle starts from origin. Then initial acceleration and velocity are

A particle starts from rest at $x=0\; m$ with an acceleration of $1 \,m / s ^2$. At $t = 5\;s _{ s }$ it receives an additional acceleration in the same direction as its motion. At $t =10\; s$ its speed and position are $v$ and $x$, respectively. Had the additional acceleration not been provided, its speed and position would have been $v _0$ and $x _0$, respectively. It is found that $x - x _0$ is $12.5 \,m$. Then one can conclude that $v - v _0$ is .............. $\,m / s$