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 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.

Answer 1

Answer:

$\frac{x_{o}}{v_B-V_A} =t$

Explanation:

Represent the car's position as a function

$x_o=$ "head start"

$x_{A}(t) = x_{o} + v_{A}t\\x_B(t)=v_Bt$

$Remember: v_B>v_A$

"cathching up means" that $x_A(t)=x_B(t)$

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