The point at which 2 vehicles meet

In summary, car A stops a traffic light, once it starts moving again it is located 330 m from car B which is traveling in the opposite direction at a constant velocity of 18 m/s. If car A accelerates at a rate of 3 m/s^2 where and after how long will the two cars pass each other?
  • #1
Hannah Wallace
13
0

Homework Statement


Car A stopped a traffic light, once it starts moving again it is located 330 m from car B which is traveling in the opposite direction at a constant velocity of 18 m/s. If car A accelerates at a rate of 3 m/s^2 where and after how long will the two cars pass each other

Homework Equations


I'm not actually sure but I think one of these equations might help me
1. Vf=vi+aΔt
2.Δd/Δt=(vi+vf)/2
3.Δd=viΔt+1/2aΔt^2
4.vf^2=vi^2+2aΔd[/B]

The Attempt at a Solution


I really don't have any idea what to do, I know that someone you need to find the point at which they interesect but I'm not sure how you arrive at that answer
 
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  • #2
Can you write separate equations of motion for each car?
 
  • #3
gneill said:
Can you write separate equations of motion for each car?
I don't know the separate equations, that's why I am asking for help
 
  • #4
Hannah Wallace said:
I don't know the separate equations, that's why I am asking for help
Consider the first car (car A). What information given in the problem statement applies to it? Which of the equation you've listed include corresponding terms?
 
  • #5
gneill said:
Consider the first car (car A). What information given in the problem statement applies to it? Which of the equation you've listed include corresponding terms?

None of them really.Maybe equation 3?
 
  • #6
Hannah Wallace said:
None of them really.Maybe equation 3?
Equation 3 looks promising. What given information applies to the terms in that equation? (First perhaps you could state in words what the terms of the equation mean, so you know what you're looking for from the problem statement to fill in the values)
 
  • #7
Well you know the distance separating the two vehicles is 330 m, you know the acceleration, you know the initial velocity which is 0, and with that you can maybe find the time separating car a from b?
 
  • #8
Ignore the second car entirely for now. Just use the information pertaining to the first car to write its equation.
 
  • #9
gneill said:
Ignore the second car entirely for now. Just use the information pertaining to the first car to write its equation.

Well I'm not sure how I could use equation 3, without taking into regards car b. what would be the distance be?
 
  • #10
Hannah Wallace said:
Well I'm not sure how I could use equation 3, without taking into regards car b. what would be the distance be?
The distance is what you're trying to find, so it is currently an unknown quantity. You want to write an equation that expresses the distance the car travels with respect to time.
 
  • #11
gneill said:
The distance is what you're trying to find, so it is currently an unknown quantity. You want to write an equation that expresses the distance the car travels with respect to time.
I really don't understand :/
 
  • #12
Car A starts at rest (zero initial speed) and accelerates at the rate 3 m/s2. How far has it traveled after a time t?
 
  • #13
gneill said:
Car A starts at rest (zero initial speed) and accelerates at the rate 3 m/s2. How far has it traveled after a time t?
would it be 1.5 x^2= distance?
 
  • #14
Hannah Wallace said:
would it be 1.5 x^2= distance?
Maybe... what does the x represent in that equation?
 
  • #15
gneill said:
Maybe... what does the x represent in that equation?
time?
 
  • #16
Hannah Wallace said:
time?
Right. So using more standard variable names ( x for distance, t for time) you have for car A:

x = 1.5 t2

Now take a look at the other car (again separately, this time ignoring car A). How would you describe its motion?
 
  • #17
gneill said:
Right. So using more standard variable names ( x for distance, t for time) you have for car A:

x = 1.5 t2

Now take a look at the other car (again separately, this time ignoring car A). How would you describe its motion?
well since it's at a constant velocity wouldn't it be 18=Δd/Δt ?
 
  • #18
Hannah Wallace said:
well since it's at a constant velocity wouldn't it be 18=Δd/Δt ?
Well that describes car B's speed. But it doesn't take into account the direction its heading or where it's starting from.

Okay, a couple of things to keep in mind. First, car B has an initial position that's different from car A's position. If we set our origin for measuring distance at the stoplight, car B's initial position with respect to that origin is at x = 330 (meters). The second thing is that car B is traveling towards the stoplight ("car B which is traveling in the opposite direction"). So what sign should you give its velocity?
 
  • #19
gneill said:
Well that describes car B's speed. But it doesn't take into account the direction its heading or where it's starting from.

Okay, a couple of things to keep in mind. First, car B has an initial position that's different from car A's position. If we set our origin for measuring distance at the stoplight, car B's initial position with respect to that origin is at x = 330 (meters). The second thing is that car B is traveling towards the stoplight ("car B which is traveling in the opposite direction"). So what sign should you give its velocity?
It'd be negative right?
 
  • #20
Hannah Wallace said:
It'd be negative right?
Correct.

Let me give you the full version of the kinematic equation for motion that takes into account initial position, initial velocity, and constant acceleration. It's the "all the bells and whistles" version of your third equation:

##x = x_o + v_i t + \frac{1}{2} a t^2##

where:
##~~~~x_o## is the initial position
##~~~~v_i## is the initial velocity
##~~~~a~## is the acceleration

Most of the equations of motion that you need to write will come from this equation. For car A, the initial position was zero as was the initial velocity, so those terms "disappear" and you are left with ##x = \frac{1}{2} a t^2## which you used.

For car B, just identify the values for the terms and fill out the equation. Is there an initial position? Yes: 300 meters. Is there an initial velocity? Yes: -18 m/s. Is there an acceleration? No: the velocity is constant, so a = 0 for car B. So write the equation.
 
  • #21
gneill said:
Correct.

Let me give you the full version of the kinematic equation for motion that takes into account initial position, initial velocity, and constant acceleration. It's the "all the bells and whistles" version of your third equation:

##x = x_o + v_i t + \frac{1}{2} a t^2##

where:
##~~~~x_o## is the initial position
##~~~~v_i## is the initial velocity
##~~~~a~## is the acceleration

Most of the equations of motion that you need to write will come from this equation. For car A, the initial position was zero as was the initial velocity, so those terms "disappear" and you are left with ##x = \frac{1}{2} a t^2## which you used.

For car B, just identify the values for the terms and fill out the equation. Is there an initial position? Yes: 300 meters. Is there an initial velocity? Yes: -18 m/s. Is there an acceleration? No: the velocity is constant, so a = 0 for car B. So write the equation.

So x=300t-18?
 
  • #22
gneill said:
Correct.

Let me give you the full version of the kinematic equation for motion that takes into account initial position, initial velocity, and constant acceleration. It's the "all the bells and whistles" version of your third equation:

##x = x_o + v_i t + \frac{1}{2} a t^2##

where:
##~~~~x_o## is the initial position
##~~~~v_i## is the initial velocity
##~~~~a~## is the acceleration

Most of the equations of motion that you need to write will come from this equation. For car A, the initial position was zero as was the initial velocity, so those terms "disappear" and you are left with ##x = \frac{1}{2} a t^2## which you used.

For car B, just identify the values for the terms and fill out the equation. Is there an initial position? Yes: 300 meters. Is there an initial velocity? Yes: -18 m/s. Is there an acceleration? No: the velocity is constant, so a = 0 for car B. So write the equation.

Then would you make the two equations equal to each other to find the distance?
 
  • #23
Hannah Wallace said:
So x=300t-18?
You've swapped the roles of the initial position and initial velocity (remember that velocity multiplied by time gives a distance, and the 300 m is already a distance). But otherwise, you've got the right idea.

Hannah Wallace said:
Then would you make the two equations equal to each other to find the distance?
Well, you have two equations in two unknowns (distance and time). Both equations give distance as a function of time, so if you set them equal to each other the unknown that's left is time. That will be the time that their positions are the same. Once you have the time you can find the position from either of your equations of motion.
 

Related to The point at which 2 vehicles meet

1. What is the point at which 2 vehicles meet?

The point at which 2 vehicles meet is known as the collision point or the point of impact. It is the exact location where the two vehicles come into contact with each other.

2. How is the point at which 2 vehicles meet determined?

The point at which 2 vehicles meet is typically determined by analyzing the damage to each vehicle and taking measurements of the vehicles' positions and orientations after the collision. This information is then used to calculate the point of impact.

3. Why is it important to determine the point at which 2 vehicles meet?

Determining the point at which 2 vehicles meet is important for accident reconstruction and determining fault in a car accident. It can also provide valuable information for improving vehicle safety and preventing future accidents.

4. Are there any factors that can affect the accuracy of determining the point at which 2 vehicles meet?

Yes, there are several factors that can affect the accuracy of determining the point at which 2 vehicles meet. These include the speed and direction of the vehicles, the type of collision (head-on, rear-end, etc.), and any external factors such as road conditions or weather.

5. Can the point at which 2 vehicles meet be determined without physical evidence?

In some cases, the point at which 2 vehicles meet can be determined without physical evidence. This can be done through eyewitness accounts, video footage, or computer simulations. However, physical evidence is typically the most accurate and reliable way to determine the point of impact.

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