physics
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Cars A and B are racing each other along a straight path in following manner: Car A has head start and is a distance dA beyond starting line at t=0. The starting line is at x=o. Car A travels at constant speed vA. Car B starts at starting line but has better engine than car A and travels at constant speed vB (which is greater than vA). How long after Car B started the race will Car B catch up with Car A? How far from Car B's starting line will the cars be when Car B passes Car A?

To summarize:
At t=0
car A: x=dA, speed=vA
car B: x=0, speed=vB
When will car B overtake car A and where.
Distance to catchup = dA
difference in speed = (vBvA)
Time to catch up, T = dA/(vBvA)
location where the two cars are sidebyside
= T*vB
= dA*vB/(vBvA) 
sdjg

Da/(VbVa)
Is the right answer.
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