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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 DA beyond the starting line at t=0. The starting line is at x=0. Car A travels at a constant speed A. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed VB, which is greater than VA.
How long after Car B started the race will Car B catch up with Car A?
Express the time in terms of given quantities.

Respuesta :

Answer:

[tex]\frac{x_{o}}{v_B-V_A} =t[/tex]

Explanation:

Represent the car's position as a function

[tex]x_o=[/tex] "head start"

[tex]x_{A}(t) = x_{o} + v_{A}t\\x_B(t)=v_Bt \\[/tex]

[tex]Remember: v_B>v_A[/tex]

"cathching up means" that [tex]x_A(t)=x_B(t)[/tex]

[tex]x_{o} + v_{A}t =v_Bt\\x_{o} = v_Bt -v_{A}t\\x_{o} = t(v_B-V_A)\\\frac{x_{o}}{v_B-V_A} =t, where \ v_B>v_A[/tex]

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