$A$ particle of mass $m$ at rest is acted upon by a force $P$ for a time $t$. Its kinetic energy after an interval $t$ is

$5\, kg$ દળ ધરાવતા એક પદાર્થ પર અચળ બળ $\overrightarrow F = {F_x}\hat i + {F_y}\hat j$ લાગે છે. $t = 0\, s$ સમયે તેનો વેગ $\overrightarrow v = (6\hat i - 2\hat j)\, m/s$ છે અને $t = 10\, s$ સમયે તેનો વેગ $\overrightarrow v = 6\hat j\, m/s$ છે. તો બળ $\overrightarrow F$ શોધો.

The linear momentum $p$ of a body of mass $5 \,kg$ varies with time $t$ as $p = 5t^2 + t + 5$. It follows that the body is moving with:

$A$ force $\vec{F}=(40 \hat{i}+10 \hat{j}) \, N$ acts on a body of mass $5 \, kg$. If the body starts from rest,its position vector $\vec{r}$ at time $t=10 \, s$ will be -

The position vector of a particle related to time $t$ is given by $\overrightarrow{r} = (10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}) \text{ m}$. The direction of the net force experienced by the particle is:

State Newton's second law of motion.