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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.
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.
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