Kinematics - Motion along a straight line

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Homework Help Overview

The discussion revolves around a kinematics problem involving two cars racing along a straight line, where one car has a head start and travels at a constant speed while the other car, starting from the same line, travels faster. Participants are attempting to determine the time it takes for the faster car to catch up to the slower car, expressed in terms of given quantities.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up equations based on the positions of the cars over time, equating their positions to find the time of intersection. Some participants express confusion about their algebraic manipulations and seek clarification on their reasoning.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with some participants questioning their own equations and others suggesting the use of relative velocity to simplify the analysis. A participant claims to have found the answer, but this has not been universally confirmed.

Contextual Notes

Some participants express uncertainty about their derived equations and whether they are correct, indicating a lack of consensus on the interpretations and methods being used.

SA32
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I’m having difficulty with an assignment for my Physics class.

Here’s the question:

“Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].

How long after Car B started the race will Car B catch up with Car A?
Express the time in terms of given quantities.”

My embarrassingly unsuccessful attempt:
My thinking was that, when Car B catches Car A, [tex]x_{A} = x_{B}[/tex].

I substituted the given quantities for Car A and Car B separately into the equation,
[tex]x = x_{o} + v_{ox}(t-t_{o}) + \frac{1} {2}*a_{x}(t-t_{o})^2[/tex]

where:
x = position as a function of time
xo = initial position
vox = initial velocity
t = a certain time
to = initial time
ax = acceleration

For Car A, I got (simplified):
[tex]x = D_{A} + v_{A}(t)[/tex]

For Car B, I got (also simplified):
[tex]x = v_{B}(t)[/tex]

Equating them, [tex]D_{A} + v_{A}(t) = v_{B}(t)[/tex], and solving for t, I got:
[tex]\frac{v_{B}t - D_{A}} {v_{A}} = t[/tex]

Which is incorrect.

I know I shouldn’t have “t” on the left side, but other than that I’m completely lost and would really appreciate it if anyone could point me in the right direction.
 
Last edited:
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Having issues fixing the fraction in my answer, so I'll try again here.

Here is what I got:

[tex]\frac{v_{B}t - D_{A}} {v_{A}} = t[/tex]
 
It may be easier to consider their relative velocity ([itex]v_{r} = v_{B}-v_{A}[/itex]).
 
Thanks a bunch, got the answer!
 
I have the same question with the cars. I've made car A equal to car B and was able to get this equation:

Da + va(t) = vb(t) and then i subtracted va(t) and then took out the common factor of t like so:

Da= t(vb-va)

then found that t= Da/(vb-va)

I'm not sure if this is the right equation, I'm pretty sure it's not so if anyone can help me that would be great!
 
sunbunny said:
I have the same question with the cars. I've made car A equal to car B and was able to get this equation:

Da + va(t) = vb(t) and then i subtracted va(t) and then took out the common factor of t like so:

Da= t(vb-va)

then found that t= Da/(vb-va)

I'm not sure if this is the right equation, I'm pretty sure it's not so if anyone can help me that would be great!
What makes you think your answer is incorrect? If your question is the same as the one originally asked then your solution is correct.
 

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