How Far Will Car A Travel Before Being Overtaken by Car B?

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In summary: Can you show me your work so I can see where I went wrong?In summary, the conversation revolves around solving a kinematics problem involving two cars traveling at different speeds on a straight road. The person asking for help has graphed the problem and is considering using the kinematic equation x_{f}=x_{i}+v_{0}t+\frac{1}{2} a t^2, but is unsure how to equate the final positions of the two cars. The expert suggests finding the intersection point on the graph, which represents the time at which the two cars come together. The expert also clarifies that the kinematic equations and the graph are essentially the same thing. The person asking for help then solves for
  • #1
prace
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Hello,

I am having an issue with this simple kinematics problem. I have the question visualized, well, at least I think I do. Here is the question:

Two cars are traveling along a straight road. Car A maintains a constand speed of 80 km/h; car B maintains a constant speed of 110 km/h. At t = 0, car B is 45 km behind car A. How much farther will car A travel before it is overtaken by car B?

So to start off, I graphed the problem and came up with this:

http://album6.snapandshare.com/3936/45466/853951.jpg
where series 1 is car A and series 2 is car B.

I also know that this has to do with position and time, so I am most likely going to use one of my kinematical equations that deals with position and time, so I chose the equation: [tex]x_{f}=x_{i}+v_{0}t+\frac{1}{2} a t^2[/tex] because I know that the accelleration is constant and is equal to 0 since the cars are not speeding up. I also know that the final positions have to be equal, but how do I equate the final positions of the two cars? Or is this even the right direction for me to solve the problem?

I mean, I could just graph the two lines, find the equations, and then find the intersection of those two lines, but I would like to learn how to do it using the kinematical equations.

Thanks
 
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  • #2
You are on the right track. By finding the intersection point on that graph you have calculated how much time passes before the two cars come together. Now you must just realize that the problem is just asking how far does car A travel during this time.

The 2 lines that you graphed are the exact same thing as the kinematic equations btw. You probably graphed something like y = 80x vs y = 110x -45.

Now look at the kinematic equation

xf = xi + vit + 1/2at^2
you stated that acceleration is 0 so its just
xf = xi + vit

final position is just the independent variable (y). xi is the initial position (y-intercept). vi is the initial velocity in the problem (coefficient of x), and T is the variable (x).

Once you set up the 2 equations you just find the intersection point by setting the 2 equations equal to each other. then you can solve for the time
 
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  • #3
So that's it? Just do it mathmatically and it is fine? Awesome, I solved for the intersection time to be .6 hours and then plugged that back into my equation for Car A and got a distance of 4.8 km. Does that sound good?

Thank you for your help Bishop!
 
  • #4
I'm currently working on this problem and I agree with that I need to find equations for both cars and then set them equal to each other. However, I did not get .6 hrs for the time. I ended up with 1.5 hours.
 
  • #5
for your help!

Dear student,

Thank you for reaching out about this kinematics problem. It seems like you have a good understanding of the problem and have made some good progress in your approach. Let me guide you through the rest of the steps to solve this problem using the kinematic equations.

First, let's define some variables to make things easier to understand:

x_A = position of car A
x_B = position of car B
v_A = velocity of car A
v_B = velocity of car B
t = time

Since we are looking for the distance traveled by car A before it is overtaken by car B, we can set up the following equation:

x_A + distance traveled by car A = x_B + distance traveled by car B

Since we know that car A and car B have constant velocities, we can use the formula for average velocity (v = d/t) to express the distance traveled by each car:

distance traveled by car A = v_A * t
distance traveled by car B = v_B * t

Substituting these into our original equation, we get:

x_A + v_A * t = x_B + v_B * t

Now, we can use the given information to solve for the time (t) when car B catches up to car A. We know that at t = 0, car B is 45 km behind car A. So, we can set up the following equation:

x_B = x_A + 45 km

Substituting this into our previous equation, we get:

x_A + v_A * t = x_A + 45 km + v_B * t

Simplifying, we get:

(v_B - v_A) * t = 45 km

Now, we can solve for t:

t = 45 km / (v_B - v_A)

Finally, to find the distance traveled by car A before it is overtaken by car B, we can use the formula for distance (d = v * t):

distance traveled by car A = v_A * t = v_A * (45 km / (v_B - v_A))

I hope this helps you understand how to use the kinematic equations to solve this problem. Keep up the good work and happy problem-solving!
 

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. It focuses on the position, velocity, and acceleration of objects in motion.

2. What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both the speed and direction of an object's motion.

3. How is acceleration calculated?

Acceleration is calculated by dividing the change in an object's velocity by the time it takes for that change to occur. It is measured in meters per second squared (m/s²).

4. What is the difference between average and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time taken, while instantaneous velocity is the velocity of an object at a specific moment in time.

5. How does the displacement-time graph relate to an object's motion?

The displacement-time graph shows the change in an object's position over time. The slope of the graph represents the object's velocity, and the area under the graph represents the object's displacement.

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