The point at which 2 vehicles meet

[Hannah Wallace]

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

 

[gneill]

Can you write separate equations of motion for each car?

 

[Hannah Wallace]

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

 

[gneill]

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?

 

[Hannah Wallace]

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?

 

[gneill]

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)

 

[Hannah Wallace]

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?

 

[gneill]

Ignore the second car entirely for now. Just use the information pertaining to the first car to write its equation.

 

[Hannah Wallace]

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?

 

[gneill]

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.

 

[Hannah Wallace]

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 :/

 

[gneill]

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?

 

[Hannah Wallace]

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?

 

[gneill]

would it be 1.5 x^2= distance?

Maybe... what does the x represent in that equation?

 

[Hannah Wallace]

Maybe... what does the x represent in that equation?

time?

 

[gneill]

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?

 

[Hannah Wallace]

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 ?

 

[gneill]

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?

 

[Hannah Wallace]

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?

 

[gneill]

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.

 

[Hannah Wallace]

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?

 

[Hannah Wallace]

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?

 

[gneill]

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.

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.