YGSK
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Hello - please can someone explain how you work this out. I have the mark scheme here but I don’t really understand how they got that formula in terms of t for the last question. Many thanks
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user342
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(Original post by YGSK)
Hello - please can someone explain how you work this out. I have the mark scheme here but I don’t really understand how they got that formula in terms of t for the last question. Many thanks
Well when A overtakes B, they've both travelled the same distance since they started from the same position. So they are equating the distances they've travelled and solving this equation for t. The distance is the area under the graph.
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mqb2766
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Note - a bluffers way would be to double the time it takes until the have the same velocity (crossing point), or equivalently, the time takes to get to 18.
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(Original post by YGSK)
Hello - please can someone explain how you work this out. I have the mark scheme here but I don’t really understand how they got that formula in terms of t for the last question. Many thanks
Another thing to realise is that A overtakes B while it is still accelerating, so they are using the formula s=ut+1/2at^2 for the displacement of A.
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