Solving 1D Motion Problems with Fun: Average Rate and Time Calculations

In summary, the problem involves a figure where a stream of people are trying to escape through a closed door. The people move towards the door at a speed of 3.50 m/s, each with a depth of 0.25 m and separated by 1.75 m. The question is asking for the average rate at which the layer of people at the door increases, and at what time the layer's depth reaches 5 m. The problem is in the 1D motion section of fundamentals of physics. To solve it, the distance to the door and the speed of the person closest to it should be used to calculate the time it takes for the person to reach the door.
  • #1
frasifrasi
276
0
FUN 1d motion prob--help

Homework Statement



The figure (http://img471.imageshack.us/my.php?image=picprobfd5.png) shows a general situation where a stream of people attempt to escape through a door that is closed. The people move toward the door at a speed of v = 3.50 m/s, are each d = .25 m in depth, and are separated by L = 1.75m. The arrangement in the figure occurs at t = 0. a.) at what average rate does the layer of people at the door increase? b.) at what time does the layer's depth reach 5 m ?

Homework Equations



The picture I drew is exactly like the one in the book.

The Attempt at a Solution



I relly have no idea how to start, but it is the 1d motion section of fundamentals of p.

Thank you.
 
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  • #2
Pleaseeeee
 
  • #3
How long does it take for the guy closest to the door to get to the door... The length of the layer increases from 0 to 0.25m, in this amount of time...
 
  • #4
learningphysics, all the info is there--I copied it exaclty. I really need help getting started.
 
  • #5
frasifrasi said:
learningphysics, all the info is there--I copied it exaclty. I really need help getting started.

I know... I'm trying to guide you along. :wink:

How long does it take the guy closest to the door to get to the door? Calculate this time... Look at the picture... what is the distance to the door? How fast is he walking? You have everything you need to get the time it takes...
 

Related to Solving 1D Motion Problems with Fun: Average Rate and Time Calculations

1. What is "1d motion" in physics?

"1d motion" refers to motion that occurs in a straight line, also known as one-dimensional motion. In physics, motion is described as the change in an object's position over time, and one-dimensional motion specifically refers to motion that occurs along a single axis or direction.

2. How can I solve "FUN 1d motion prob-help" problems?

To solve "FUN 1d motion prob-help" problems, you will need to use the basic equation of motion, which is displacement (Δx) = velocity (v) x time (t). You will also need to identify the known values and the unknown value in the problem and use algebraic manipulation to solve for the unknown value.

3. What is the difference between distance and displacement in 1d motion?

Distance refers to the total length or path traveled by an object, while displacement refers to the change in position of an object from its initial position to its final position. Distance is a scalar quantity, while displacement is a vector quantity that includes magnitude (distance) and direction.

4. How can I calculate average velocity in 1d motion?

To calculate average velocity in 1d motion, you will need to divide the total displacement (change in position) by the total time taken. The formula for average velocity is v = Δx / t. Remember to use the same units for both displacement and time, such as meters and seconds.

5. Can you provide an example of a "FUN 1d motion prob-help" problem?

Sure! Here's an example: A car travels at a constant speed of 30 m/s for 10 seconds. What is the car's displacement during this time? To solve this problem, we will use the formula Δx = v x t, where v = 30 m/s and t = 10 seconds. Thus, the displacement is Δx = 30 m/s x 10 s = 300 meters.

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