Paper boy problem (relative motion)

In summary, 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 front 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. The horizontal velocity of the paper with respect to the sidewalk is 7.2 m/s. To determine how far in advance of the walkway he should release the papers so that they land on the porch, the boy needs to consider the time it takes for the paper to reach the porch. This can be calculated by dividing the distance (15 m) by the horizontal velocity (6 m/s),
  • #1
DB
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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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  • #2
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
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
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...
 
  • #5
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
 
  • #6
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
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
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...
 
  • #9
ahhh, i see, thanks for the help light grav, apreciate it
 

Related to Paper boy problem (relative motion)

What is the "Paper Boy Problem" in relative motion?

The "Paper Boy Problem" is a classic physics problem that involves a person moving at a constant speed on a bicycle, throwing newspapers onto doorsteps while passing them. It is used to illustrate relative motion, which is the movement of an object in relation to another object.

What are the key concepts involved in solving the "Paper Boy Problem"?

The key concepts involved in solving the "Paper Boy Problem" include velocity, acceleration, and the concept of frames of reference. These concepts help us understand how the paper boy's speed and direction are affected by his own motion and the motion of the houses he is passing.

How do you calculate the relative velocity in the "Paper Boy Problem"?

The relative velocity in the "Paper Boy Problem" can be calculated by subtracting the velocity of the houses from the velocity of the paper boy. This gives us the velocity of the paper boy in relation to the houses, which is the relative velocity.

What is the significance of the "Paper Boy Problem" in real-life scenarios?

The "Paper Boy Problem" has real-life applications in fields such as transportation and aviation. It helps us understand how two objects moving at different speeds and directions can affect each other's motion. This is crucial for designing efficient and safe transportation systems.

What are some common misconceptions about the "Paper Boy Problem"?

One common misconception about the "Paper Boy Problem" is that the paper boy's velocity is the same as the velocity of the houses. This is not true as the paper boy is moving in relation to the houses, and his velocity is affected by his own motion. Another misconception is that the paper boy's acceleration is constant, when in fact it changes as he throws newspapers at different doorsteps.

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