Kinematics Car velocity Problem

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Homework Help Overview

The problem involves two cars moving towards each other along a straight path, with the red car's velocity given and the task of determining the initial velocity and acceleration of the green car based on their positions at the time they pass each other.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the equations used to relate the cars' positions and velocities, questioning the origin of the numbers in the equations. There is an exploration of how to check the results by substituting back into the equations.

Discussion Status

Some participants have offered suggestions on how to verify the calculations and check for missing units. There is an ongoing discussion about the signs of the velocity and acceleration, with participants seeking clarification on how to determine the appropriate signs based on the direction of motion.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to consider signs for velocity and acceleration based on the defined directions in the problem statement.

hiineko
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Homework Statement


In the figure here, a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 223 m. If the red car has a constant velocity of 20.0 km/h, the cars pass each other at x = 43.1 m. On the other hand, if the red car has a constant velocity of 40.0 km/h, they pass each other atx = 76.9 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.

Homework Equations


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The Attempt at a Solution


eq 1 = 179.9=Vo(7.75)+1/2a(7.75)^2
eq 2 = 146.1=Vo(6.92)+1/2a(6.92)^2

insert 2 to 1

Vo = 146.1-1/2a(6.92)/(6.92)

179.9=(146.1-1/2a(6.92)/(6.92))(7.75)+1/2a(7.75)^2

got a = 5.06 m/s^2
and vo as 3.60

is my answer right?
 
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You could explain where all those numbers in equation 1 and 2 come from, that makes it easier to understand the solution.
Did you plug the result back into those equations to check?
 
mfb said:
You could explain where all those numbers in equation 1 and 2 come from, that makes it easier to understand the solution.
Did you plug the result back into those equations to check?

Vr = 20km/h = 5.56m/s x = 43.1
v=d/t -> t=d/v = 43.1/5.56=7.75s
xgreencar - x = 223 - 43.1 = 179.9

179.9=vo(7.75)+1/2a(7.75)^2

same procedure to eq. 2

Sorry but I need a bit help here
*is my answer right
*if not, can you please help me?
 
I suggested a way you can test your answer yourself.

Yes it is right. Well, several units are missing.

Edit: oh right, forgot about the signs.
 
Last edited:
Check signs per original problem statement.
 
mfb said:
I suggested a way you can test your answer yourself.

Yes it is right. Well, several units are missing.

Edit: oh right, forgot about the signs.

That's one thing that bothers me I need signs how do I know what signs will I use? I just got a positive does it mean I'll use + sign?
 
hiineko said:
That's one thing that bothers me I need signs how do I know what signs will I use? I just got a positive does it mean I'll use + sign?
On the X axis,which way is the green car going?
 
insightful said:
On the X axis,which way is the green car going?

So 5.06 and 3.61 will be put a - sign on it? If that's what you meant
 
Correct, if velocity and acceleration are in the (defined) negative direction, they will have negative values.
 
  • #10
insightful said:
Correct, if velocity and acceleration are in the (defined) negative direction, they will have negative values.

Okay thank you! Off to my next question in my next post see you there lol kidding Thankyouu
 

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