Question

A newspaper delivery boy is riding his bicycle down the street at $5.0\mathrm{m}/\mathrm{s}$. He can throw a paper at a speed of $8.0\mathrm{m}/\mathrm{s}$. What

is the paper’s speed relative to the ground if he throws the paper (a) forward, (b) backward, and (c) to the side?

Short answer

a. speed is $13\mathrm{m}/\mathrm{s}$

b. speed is $-3\mathrm{m}/\mathrm{s}$

c. Speed is$9.43\mathrm{m}/\mathrm{s}.$

Step-by-step solution

Party (a) Step 1: Given information

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.

Simplify

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,

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

Part (b) Step 1: Give information

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.

Simplify

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

then,

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

Part (c) Step 1: Given information

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.

Simplify

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,

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