Paper boy problem (relative motion)

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Homework Help Overview

The discussion revolves around a relative motion problem involving a newspaper boy throwing papers while riding his bicycle. The scenario includes calculating the horizontal velocity of the papers relative to the sidewalk, determining the distance to release the papers for them to land on the porch, and finding the angle at which to throw the papers when directly opposite the porch.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of the horizontal velocity of the papers and discuss the implications of time in relation to the motion of the paper and the boy. There are questions about how to approach the distance for releasing the papers and the angle needed for the throw.

Discussion Status

Some participants have provided insights into the relationship between time and motion, suggesting that the time it takes for the paper to reach the porch is crucial for determining the release point. Others have attempted calculations but express uncertainty about their results, indicating a mix of understanding and confusion regarding the problem.

Contextual Notes

Participants note that the problem does not provide explicit formulas or methods, leading to questions about how to approach the calculations. There is also mention of the need to keep perpendicular velocities separate while considering their independence.

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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

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

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

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

okay here where i don't 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
 
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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.
 
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.
 
i know that this is probably a simple question, but still can't get it, I've got that it will take 2.1 secs for the paper to reach the porch but i feel like I am missing something...
 
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 isn't the answer, the answer to b is 10 btw
 
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.
 
i understand the fact that perpendicular velocities are independent.
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?
 
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:
|------
|
|
|

4[m/s]/6[m/s] = y/15[m]

location triangle:

|-------------
|
|
|
|
|
|
|
|
|
|

(c) is easy IF you keep the components separate...
 
ahhh, i see, thanks for the help light grav, apreciate it
 

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