At $t=0$ force $F=ct$ is applied to a small body of mass $m$ resting on a smooth horizontal plane ($c$ is a constant). The force is at an angle $\theta$ with the horizontal .The velocity of the body at the moment of its breaking off the plane and the distance travelled by the body up to this moment are 

Since $F \cos \theta =\mathrm{ma}$ or $\mathrm{ct} \cos \theta =m\frac{dv}{dt}$

${\int }_0^{v}dv=\frac{c \cos \theta }{m}{\int }_0^{t}t dt$ or ${\int }_0^{t}t dt or v=\left(\frac{c \cos \theta }{2m}\right) t^2$

But velocity at the time of breaking off or at

$$\begin{gathered} t\mathit{ }\mathit{=}\frac{mg}{c\mathit{ }sin\mathit{ }\theta } \\ {\mathit{.}}_{\mathit{ }\mathit{ }}^{\mathit{.}}\mathit{.}\mathit{ }v\mathit{=}\frac{c\mathit{ }cos\mathit{ }\theta }{\mathit{2}m}x\mathit{ }{\mathit{\left(}\frac{mg}{c\mathit{ }sin\mathit{ }\theta }\mathit{\right)}}^{\mathit{2}}\mathit{=}\frac{m{g}^{\mathit{2}}cos\mathit{ }\theta }{\mathit{2}c\mathit{ }si{n}^{\mathit{2}}\mathit{ }\theta } \\ Now\mathit{,}\mathit{ }v\mathit{=}\frac{dS}{dt}\mathit{=}\frac{c\mathit{ }cos\mathit{ }\theta }{\mathit{2}m}{t}^{\mathit{2}} \\ {\int }_0^{s}dS\mathit{=}\frac{c\mathit{ }cos\mathit{ }\theta }{\mathit{2}m}{\mathit{\int }}_{\mathit{0}}^t\mathit{ }tt\mathit{ }dt\mathit{ }or\mathit{ }S\mathit{=}\mathit{\left(}\frac{c\mathit{ }cos\mathit{ }\theta }{\mathit{6}m}\mathit{\right)}{t}^{\mathit{3}} \end{gathered}$$

Now, $v\mathit{=}\frac{dS}{dt}\mathit{=}\frac{c\mathit{ }cos\mathit{ }\theta }{\mathit{2}m}{t}^{\mathit{2}}$

${\int }_0^{s}dS\mathit{=}\frac{c\mathit{ }cos\mathit{ }\theta }{\mathit{2}m}{\mathit{\int }}_{\mathit{0}}^t\mathit{ }t\mathit{ }dt$ or $S\mathit{=}\mathit{\left(}\frac{c\mathit{ }cos\mathit{ }\theta }{\mathit{6}m}\mathit{\right)}{t}^{\mathit{3}}$