# 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?

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1. Da/(Vb-Va)
Is the right answer.

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2. that helped me get a full mark thanks

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3. A test car travels in a straight line along the x axis. The graph in the figure shows the car's position x as a function of time. Find a) the average velocity between t₁ = 0.0 s and t₂ = 6.0 s. b) the instantaneous velocity at points A, D and G. c) The instantaneous speed at points A, D and G. d) the average speed between t₁ = : 0.0 s and t₂ = 6.0 s

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4. 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 catch-up = dA
difference in speed = (vB-vA)
Time to catch up, T = dA/(vB-vA)
location where the two cars are side-by-side
= T*vB
= dA*vB/(vB-vA)

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