How can you tell if the interaction is isotropic?

In summary, the teenager drove for 5 minutes at a speed of 18 mph and then came to a stop. His average speed for the trip was 13.9 mph.
  • #1
FalconKICK
30
0

Homework Statement


A teenager has a car that accelerates at 3.00 m/s2 and decelerates at -4.50 m/s2. On a trip to the store, he accelerates from rest to 16.5 m/s, drives at a constant speed for 5.00 s, and then comes to a momentary stop at the corner. He then accelerates to 18.0 m/s, drives at a constant speed for 15.0 s, and decelerates for 2.67 s. Then he continues for 4.00 s at this speed and comes to a stop.


Homework Equations


(a) How long does the trip take?
(b) How far has he traveled?
(c) What is his average speed for the trip?
(d) How long would it take to walk to the store and back if he walks at 1.50 m/s?


The Attempt at a Solution


I don't know which equations and what numbers to use. Any help on how I can find out which numbers are the ones I need and which equation I should use?
 
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  • #2
Get some graph paper and draw a diagram.
Put time along the bottom and speed up the side.
The area under the graph is the distance.

Then when you understand it you can use s=ut + 1/2 at^2 to solve each section.
 
  • #3
mgb_phys said:
Get some graph paper and draw a diagram.
Put time along the bottom and speed up the side.
The area under the graph is the distance.

Then when you understand it you can use s=ut + 1/2 at^2 to solve each section.

I've actually done that but I'm still confused.
 
  • #4
You should be able to easily plot the speed at each point in time - it's just a series of straight lines.

A teenager has a car that accelerates at 3.00 m/s2 and decelerates at -4.50 m/s2.
So we assume all accel is 3m/s^2 and all braking is -4.5m/s^2 (actually if you are going to say deceleration and have a negative vallue that cancels out and implies aceleration ...)

he accelerates from rest to 16.5 m/s,
Use v = u + a t with 3.0 m/s^2 to get 't'


drives at a constant speed for 5.00 s,
Easy

and then comes to a momentary stop at the corner.
Use v=u + a t again with -4.5m/s^2

He then accelerates to 18.0 m/s
Use v = u + a t with 3.0 m/s^2 to get 't'

drives at a constant speed for 15.0 s,
Easy

and decelerates for 2.67 s.
Use v=u + a t again with -4.5m/s^2

Then he continues for 4.00 s at this speed
Easy

and comes to a stop.
Use v=u + a t again with -4.5m/s^2

Then the time you can just read of the axis.
The distance is the integral (ie the area under the curve)
And the average speed is just distance/time
 
  • #5
Is it the same thing as finding the X and Y components? If not I'm confused. I was never really good at physics anyways.
 
  • #6
I got the time the trip takes. All I really need now is how far he traveled, but I am having trouble figuring out how to get this. Any help?

Edit: I solved it :D
 
Last edited:

Related to How can you tell if the interaction is isotropic?

1. How is isotropy defined in science?

Isotropy is defined as a property of a system or interaction that is independent of direction or orientation. This means that the system or interaction is the same in all directions and there is no preferred direction or axis.

2. What are some examples of isotropic interactions?

Examples of isotropic interactions include the force of gravity, electromagnetic radiation, and diffusion of particles in a liquid or gas. In each of these cases, the interaction is the same in all directions.

3. How can you tell if an interaction is isotropic?

An interaction is considered isotropic if the measured quantities or properties of the system do not change when the direction or orientation is changed. This can be determined through experiments or mathematical calculations that show consistent results regardless of direction.

4. What are some techniques used to test for isotropy?

Some techniques used to test for isotropy include rotational symmetry tests, polarization measurements, and statistical analysis. These methods help to identify any patterns or biases in the data that could indicate non-isotropic behavior.

5. Why is it important to determine if an interaction is isotropic?

Determining if an interaction is isotropic is important because it helps us understand the fundamental properties of a system and can inform our understanding of natural phenomena. It also allows us to make accurate predictions and develop models that can be applied in various fields of science and technology.

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