Calculating Distance and Time for Two Cars on a One-Dimensional Road

In summary, there is a conversation about finding when two cars traveling at different speeds will meet. The first car travels at 20 m/s and passes a line at time t=0s, while the second car travels at 30 m/s and passes the same line 5 seconds later. After discussing equations and solving for t=15 seconds, there is confusion about finding the distance between the cars when they meet. It is mentioned that the book being used for reference is translated poorly.
  • #1
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An automobile which travels with the constant speed 20 m/s passes in a section in the moment t=0s and 5 seconds later in the same section passes another automobile which travles with the speed of 30 m/s in the same direction.a) Find when the second car meets the first one.
My solution : x1=x2 20t=30(t-5) and here we find t=15 seconds.b) Whats the distance between the cars since the section until they meet? How to do this one? Just give me an idea..
 
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  • #2
Hi Elaia06! :smile:

hmm :confused:

ahh! o:), i think you mean …​
Elaia06 said:
An automobile which travels with the constant speed 20 m/s passes a line at time t=0s and 5 seconds later another automobile which travels with the speed of 30 m/s in the same direction passes the same line.
a) Find when the second car meets the first one.
My solution : x1=x2 20t=30(t-5) and here we find t=15 seconds.b) Whats the distance past the line when they meet?

your equations should be …
x1 = 20t
x2 = 30(t-5)
x1 = x2 :wink:

you already have t = 15,

so x1 = x2 = … ? :smile:
 
  • #3
I did it this way, but my teacher said it was not correct o_O Now I logged in physicsforum and I see we have the same opinion :S Even the other students had done it this way.Whatever,thank you :)
 
  • #4
Hi Elaia06! :smile:

Please remember that I rewrote the question so that it fitted your solution!

Perhaps the question means something different?
 
  • #5
So the distance question means " The two cars,since they pass the line have a distance.They keep this distance until they meet.Find the distance.This is basically what I did,found that distance.
 
  • #6
Elaia06 said:
" The two cars,since they pass the line have a distance.They keep this distance until they meet.Find the distance.

sorry, but that doesn't make any sense :redface:

(how can they keep the distance if the distance is getting smaller?)
 
  • #7
Thats what it says in my book,literally.I am not english and the book I work with is translated poorly ..Whatever,thank you anyway..
 

1. What is motion in one dimension?

Motion in one dimension refers to the movement of an object along a straight line, with no change in direction. This type of motion is commonly represented by distance-time graphs, where the position of the object is plotted against time.

2. How is speed different from velocity in one dimension?

Speed is a measure of how fast an object is moving, while velocity is a measure of how fast an object is moving in a specific direction. In one dimension, speed and velocity are the same as there is no change in direction.

3. What is acceleration in one dimension?

Acceleration in one dimension is the rate of change of velocity over time. It can be positive, negative, or zero, depending on whether the object is speeding up, slowing down, or maintaining a constant velocity.

4. How do I calculate displacement in one dimension?

Displacement in one dimension is the change in an object's position from its initial position to its final position. It can be calculated by subtracting the initial position from the final position.

5. What are the equations of motion in one dimension?

The equations of motion in one dimension are:

1. v = u + at (relates initial velocity, final velocity, acceleration, and time)

2. s = ut + ½at^2 (relates displacement, initial velocity, acceleration, and time)

3. v^2 = u^2 + 2as (relates final velocity, initial velocity, acceleration, and displacement)

4. s = ½ (u + v)t (relates displacement, initial velocity, final velocity, and time)

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