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m1 vector question

d^2 = 25t^2 - 112t +169

find the length of time to the nearest minute for which the distance is less than 11km
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Avatar for steve2005
steve2005
17
Original post by examweek
d^2 = 25t^2 - 112t +169

find the length of time to the nearest minute for which the distance is less than 11km


d^2=121

Form a quadratic and factorise .
Reply 2
Avatar for raheem94
raheem94
13
Original post by examweek
d^2 = 25t^2 - 112t +169

find the length of time to the nearest minute for which the distance is less than 11km


112>25t2−112t+169  ⟹  0>25t2−112t+48  ⟹  25t2−112t+48<0 11^2 > 25t^2 -112t +169 \implies 0> 25t^2 -112t +48 \implies 25t^2 -112t +48 < 0

Now find the critical values by solving, 25t2−112t+48=0 25t^2 -112t +48 = 0

At the end draw the graph, and find the required length of time.

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