Efficiently Solve Your Car Lane Change Time Problem - 65.0 km/h vs 45.0 km/h

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Homework Help Overview

The problem involves calculating the time it takes for a car traveling at 65.0 km/h in the left lane to pull alongside another car traveling at 45.0 km/h in the right lane, given that their front bumpers are initially 100 m apart. This falls under the subject area of kinematics.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations and the implications of constant speed. There is a focus on how to set up the equations based on the relative positions and velocities of the two cars. Some participants express uncertainty about the next steps after converting units.

Discussion Status

There are multiple approaches being explored, including the use of kinematic equations and relative motion concepts. Some participants have provided guidance on how to set up the equations, while others are questioning the necessity of unit conversions and discussing different methods for solving the problem.

Contextual Notes

Participants are navigating unit conversions between km/h and m/s, and there is a mention of the need to consider both the distances and velocities in consistent units. The initial separation of 100 m is also being discussed in the context of these conversions.

motionman04
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How long does it take an automobile traveling in the left lane at 65.0 km/h to pull alongside a car traveling in the same direction in the right lane at 45.0 km/h if the cars' front bumpers are initially 100 m apart?
 
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What have you done thus far?
 
Use Kinematic equations.

Both have constant speed so

[tex]v_{1}t = x[/tex] for the car on the right lane

[tex]v_{2}t = x + 100[/tex] for the car on the left lane
 
Well I did some conversions to obtain 18.05 m/s as the velocity of the first car, but still kinda unsure of where to go from there.
 
Don't bother with conversions. You know the velocities. X for both cars is the same (that's why the second equation has the extra 100). The times will be the same. You are in a position with two equations and two unknowns. There are a handful of ways to solve this. You could use substitution, plug the coefficients into a matrix and rref it, solve simultaneously... Put all of the things you don't know on the left side with the respective coefficients. Leave the suff you do know on the right (like the 100). And solve.

Good luck.
 
but if you do not change the velocities to m/s then you must change the 100m to .1 km.

You could use relative motion...

the velocity of the left one with respect to the right is the velocity of the left minus the velocity of the right one...so 20km/hr. Then you want the relative velocity to go 100m (or .1 km)...so .1 km / 20 km/h should yield the correct answer in hours...convert to seconds by multiplying by 3600.
 

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