When will these cars pass eachother - kinematics

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Homework Help Overview

The discussion revolves around a kinematics problem involving two cars, Car A and Car B, traveling towards each other on a highway. Car A starts from rest and accelerates, while Car B travels at a constant speed and later decelerates. The goal is to determine the acceleration of Car A when both cars pass each other at a specific point.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations of motion for both cars, considering their initial conditions and accelerations. There are attempts to set up equations based on their velocities and positions at the moment they pass each other. Some participants express uncertainty about the correctness of their equations and seek clarification.

Discussion Status

The discussion includes various attempts to formulate the problem mathematically, with some participants providing equations for the motion of each car. There is a recognition of the need to solve a system of equations, but no consensus has been reached on the correct approach or solution yet.

Contextual Notes

Participants note the specific conditions of the problem, including the initial speeds and the point at which the cars pass each other. There is mention of a second part of the problem that involves finding the distance between the two cars, which adds complexity to the discussion.

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Car A with u=0/s is on the northbound land of a highway and car B is traveling in th southbound lane at constant u=26.67m/s.At t=0 , A starts and accelerates at a constant rate a(A) . while at t=5 , B decelerates with constant magnitude of a(A)/6 . knowing that when the two carspass each other at x=90m , and v(A) = v(B),determine acceleration of car A.

i have tried it many times but stil can't ind the answer.pls help.
the answer is 3.59m/s^2
thanx
 
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oh...the length between the two cars is d in which we have to find in the second part
 
You just need to write the equations of velocity and movement of each car. Then you'll find out the answer[/color]
 
for v(A)=180a(A)--------1
s(A)=1/2 at^2--------2

v(B)=26.67 - 0.5a(B)t--------3
d-90=26.67t - 0.5a(B)t^2-----4

is it rite??
 
pls help...
 
pls help...
 
teng125 said:
pls help...

Teng! I'm not a moderator but all you have to do is ask ONCE!

Tricky! I like it.

In the following, I have defined + to be Northward.
For car A: Car A starts at position [tex]x_{0A}[/tex] at a speed of [tex]v_{0A} = 0m/s[/tex] with an acceleration of a(A).
[tex]x_A=x_{0A}+v_{0A}t+(1/2)a_At^2[/tex]; [tex]v_A=v_{0A}+a_At[/tex]
[tex]x_A=x_{0A}+(1/2)a(A)t^2[/tex]; [tex]v_A=a(A)t[/tex]

For car B: Car B starts at position [tex]x_{0B}[/tex] at a speed of [tex]v_{0B} = -26.67 m/s[/tex] with an acceleration of -(1/6)a(A).
[tex]x_B=x_{0B}+v_{0B}t+(1/2)a_Bt^2[/tex]; [tex]v_B=v_{0B}+a_Bt[/tex]
[tex]x_B=x_{0B}-26.67t-(1/12)a(A)t^2[/tex]; [tex]v_A=-26.67-(1/6)a(A)t[/tex]

Now, at x=90 m, both cars pass each other and have the same speed. We don't have a specific equation to find t or a(A), but we have two equations we can write. First, at time t both cars have the same speed. Thus at (90 m, t):
[tex]a(A)t=-26.67-(1/6)a(A)t[/tex]

Second, both cars are at the same point at this time, so:
[tex]x_{0A}+(1.2)a(A)t^2=x_{0B}-26.67t-(1/12)a(A)t^2 =90 m[/tex]
(This last is actually THREE equations. Both distances are equal to 90 m as well as each other.)

You've got four equations in four unknowns, so you should be able to solve the system.

-Dan
 

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