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Cars A and B are racing each other along the same straight road in 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 VA. 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.(a) How long after Car B started the race will Car B catch up with Car A? (b) How far from Car B's starting line will the cars be when Car B passes Car A?

Respuesta :

estervarga

Answer:

tA=tB

tB=(sA-dA)/vA watch it below

Explanation:

the spatio for A is sA=vA*tA+dA

the spatio for B is sB=vB*tB

the meeting point will happen at: tA=tB

so, sA=vA*tB+dA

tB=sB/vB

yields:

sA=vA*(sB/vB)+dA

sA-dA=vA*(sB/vB)

sB=(sA-dA)*vB/vA

then

tB=((sA-dA)*vB/vA)/vB

tB=(sA-dA)/vA

The answer is tA=tB

When tB= (sA-dA) /vA watch it below

  • Then the section for A is sA=vA*tA+dA
  • After that the section for B is sB=vB*tB

When the meeting point will happen at  tA=tB

  • Now,  sA=vA*tB+dA
  • After that tB=sB/vB

After that yields:

  • Then sA= vA*(sB/vB)+dA
  • Then sA-dA= vA*(sB/vB)
  • Now, sB= (sA-dA)*vB/vA
  • After that tB=((sA-dA)*vB/vA)/vB
  • Thus, tB=(sA-dA)/vA

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