# Easy M1 Question!Watch

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#1
but i still cant do it!
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#2
I thought that if something is parallel to something, the coefficients are the same. So here i thought it would be 2q^2-3 = 1, but its not the case. The mark scheme says that : 2q^2-3 = -(q+2)..and then solve it. But why!!!
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#3
0
14 years ago
#4
If parallel, the vectors can be expressed as (ai+bj) and k(ai+bj) i.e. the ratio of the i and j coefficients are equal (one definition of parallel)
So in this case:-
2q^2-3 = -q-2
2q^2 + q - 1 = 0
(2q - 1)(q + 1) = 0
-> q = -1 or 1/2
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14 years ago
#5
q = -1
-> Velocity of Pedestrian is: (-i + j)ms^-1
r = ro + (-i+j)t
5 > (6i-j) + (-i+j)t
5 > (6-t)i + (t-1)j
5 > √((6-t)² + (t-1)²)
25 > (6-t)² + (t-1)²
t²-12t+36 + t²-2t+1 - 25 < 0
2t² -14t + 12 < 0
t² - 7t + 6 < 0
(t - 6)(t - 1) < 0
1 < t < 6
i.e. length of time = 5s
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