91Ó°ÊÓ

we have given,

$$\begin{gathered} \mathrm{Bicycle}\mathrm{speed}=5\mathrm{m}/\mathrm{s} \\ \mathrm{Paper}\mathrm{speed}=8\mathrm{m}/\mathrm{s} \end{gathered}$$

we have to find the paper relative speed with ground if he through it forward.

If boy through the paper forward then there will be speed of the bicycle ${\mathrm{v}}_{\mathrm{b}}$and the speed of the paper ${\mathrm{v}}_{\mathrm{p}}$,

Then total speed of the paper with respect to ground will add,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{g}}={\mathrm{v}}_{\mathrm{b}}+{\mathrm{v}}_{\mathrm{p}} \\ {\mathrm{v}}_{\mathrm{g}}=5\mathrm{m}/\mathrm{s}+8\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{g}}=13\mathrm{m}/\mathrm{s} \end{gathered}$$

We have given,

$$\begin{gathered} \mathrm{Bicycle}\mathrm{speed}=5\mathrm{m}/\mathrm{s} \\ \mathrm{Paper}\mathrm{speed}=8\mathrm{m}/\mathrm{s} \end{gathered}$$

We have to find the speed of paper with respect to ground if it flows back ward.

Since the paper is flying backward then its speed with the cycle will be negative.

then,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{g}}={\mathrm{v}}_{\mathrm{b}}+{\mathrm{v}}_{\mathrm{p}} \\ {\mathrm{v}}_{\mathrm{g}}=5\mathrm{m}/\mathrm{s}-8\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{g}}=-3\mathrm{m}/\mathrm{s} \end{gathered}$$

We have given,

$$\begin{gathered} \mathrm{Bicycle}\mathrm{speed}=5\mathrm{m}/\mathrm{s} \\ \mathrm{Paper}\mathrm{speed}=8\mathrm{m}/\mathrm{s} \end{gathered}$$

We have to find the speed of the paper with respect to ground if it flows to the side.

The velocity of the paper with respect to ground is,

${\mathrm{v}}_{\mathrm{g}}=5\mathrm{m}/\mathrm{s}\hat{\mathrm{i}}+8\mathrm{m}/\mathrm{s}\hat{\mathrm{j}}$

Then magnitude will be,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{g}}=\sqrt{{\left(5\right)}^2+{\left(8\right)}^2} \\ {\mathrm{v}}_{\mathrm{g}}=\sqrt{25+64}=\sqrt{89} \\ {\mathrm{v}}_{\mathrm{g}}=9.43\mathrm{m}/\mathrm{s} \end{gathered}$$