Kinematics Car velocity Problem

AI Thread Summary
The discussion revolves around a kinematics problem involving two cars moving towards each other, with the red car's velocities given as 20.0 km/h and 40.0 km/h. The user attempts to calculate the initial velocity and acceleration of the green car using equations derived from the cars' positions at the time they pass each other. They find values of acceleration as 5.06 m/s² and initial velocity as 3.60 m/s but express confusion about the signs for these values. Clarifications are provided regarding the need to assign negative signs based on the defined direction of motion on the x-axis. The user is encouraged to verify their results and understand the significance of signs in their calculations.
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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