# Paper boy problem (relative motion)

1. Nov 10, 2005

### DB

a newspaper boy throws newspapers sideways onto the porches of his customers while riding his bicycle along the sidewalk. the sidewalk is 15 m in from of the porches. the boy throws the papers at a horizontal speed of 6 m/s relative to himself and rides the bicycle at 4 m/s relative to the sidewalk.
Calculate:

a) the horizontal velocity of the paper with respect to the sidewalk. Ans: 7.2 m/s

so i made the paper boy traveling south, the papers being thrown west.
*PE means paper with respect to the earth

$$\overrightarrow{V_{PE}}=[-6,0]+[0,-4]$$

$$\overrightarrow{V_{PE}}=[-6,-4]$$

using the components a) is ~ 7.2 m/s West 34degrees South

okay here where i dont know where to start, it doesn't say how to do this in my book so i thought id ask you guys.

b)how far in advance of the walkway should he release the papers so that they land on the porch?

how would i approach this? do i just have to think hard or is some formulas that could help me solve this?

c)if he waits until he is directly opposite the porch, at what angle with respect to the sidewalk must he throw the paper so that it lands on target?

i think i'll need the answer to "b" to solve this no? if not some hints would be greatly apreciated.

thanks guys

2. Nov 10, 2005

### Pengwuino

Well for B) it is actually quite simple. You know at what angle the paper is thrown (or at least you can easily find it out with that vector) and you know how far the porch is from the sidewalk, so you should be able to determine how far north from the porch the person needs to throw it. Just draw a diagram and you should realize what the angle is.

c) This should actually be really easy since instead of approaching the porch (at say, y=0), you're simply leaving the porch at the exact same velocities.

3. Nov 10, 2005

### lightgrav

The conceptual key to part B is TIME ... how long of a time does it take
for the paper to get from the sidewalk to the porch?
We're not asked to compute this, but you MUST realize that
the paper is moving forward for that same time.

Motion in each direction (component) is independent of motion in the other directions ... they are connected to each other through TIME.

4. Nov 10, 2005

### DB

i know that this is probably a simple question, but still can't get it, ive got that it will take 2.1 secs for the paper to reach the porch but i feel like im missing something....

5. Nov 10, 2005

### DB

my thinking right now is how far does he go in 2.1 secs? but when i multiply that by 4 m/s sec it gives me 8.4 m which isnt the answer, the answer to b is 10 btw

6. Nov 10, 2005

### lightgrav

doesn't it go 15m sideways?
isn't it going 6 m/s sideways?

"Independent" means keep each direction SEPARATE!
The only thing that's the same for them is TIME.

7. Nov 10, 2005

### DB

i understand the fact that perpendicular velocities are independant.
i think that i got it maybe not the wat i should of, i used the fact that alternate interior angles are congruent with the transversal being the velocity vector and used tan to solve : tan(34)=x/15, x~10 m

an easier way?

8. Nov 10, 2005

### lightgrav

15m divided by 6 m/s = 2.5 sec.

if you have a velocity triangle, and a SIMILAR location triangle,
skip the angle! the ratio of legs is the same

velocity triangle:
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4[m/s]/6[m/s] = y/15[m]

location triangle:

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(c) is easy IF you keep the components separate...

9. Nov 10, 2005

### DB

ahhh, i see, thanks for the help light grav, apreciate it