$\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m$
The direction of net force experienced by the particle is :
Three forces $$F_{1}=10 \mathrm{~N}, F_{2}=8 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}$$ are acting on a particle of mass $$5 \mathrm{~kg}$$. The forces $$\mathrm{F}_{2}$$ and $$\mathrm{F}_{3}$$ are applied perpendicularly so that particle remains at rest. If the force $$F_{1}$$ is removed, then the acceleration of the particle is:
A body of mass $$500 \mathrm{~g}$$ moves along $$\mathrm{x}$$-axis such that it's velocity varies with displacement $$\mathrm{x}$$ according to the relation $$v=10 \sqrt{x} \mathrm{~m} / \mathrm{s}$$ the force acting on the body is:-
At any instant the velocity of a particle of mass $$500 \mathrm{~g}$$ is $$\left(2 t \hat{i}+3 t^{2} \hat{j}\right) \mathrm{ms}^{-1}$$. If the force acting on the particle at $$t=1 \mathrm{~s}$$ is $$(\hat{i}+x \hat{j}) \mathrm{N}$$. Then the value of $$x$$ will be: