some calculus anyone? see pic:

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some calculus anyone? see pic:

Mathematics
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interesting problem. I'll come back later if nobody help you :).
maybe i should write my progress too \[\tan \Theta =((4m \sin \alpha) \div (m + 4 m \cos \alpha)) +1\] for q a

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the velocity vector of the truck is 4 times the magnitude of the car then right?
add the vectors to get the new vector direction
< 0, y> <4x,4y> -------- <4x, 5y> right?
that should be
new vector is <5x,4y> then
yeah, now im with you...
whenim wring let me know lol
i had to look at the question again :)
the new angle they go off in is y over x
4y ---- = tan(t) , but how to but that in sin cos? 5x
y = sin ; and x = cos right?
yeah so \[4\sin (\alpha)/4\cos(\alpha)= \tan(\theta)\]
truck vector = <4cos(a),4sin(a)> car vecotr = <1,0>
add them to get: < 1 , 0 > <4cos(a),4sin(a)> --------------- <4cos(a) +1, 4sin(a)> = new direction
oh, my numer and denom should be reversed
which is: tan(t) = 4sin(a)/(4cos(a)+1)
b wants you to tell them the velocity, the magnitude, that this new vector is i think
try using distance formula with new vector to see
but tan =cos/sin not tan = sin/cos
scalar the inital velocity of each to u perhaps?
tan = sin/cos ; always has and always will
tan = slope = y/x = sin/cos
dam i looked at cot
bjargh all i can get is \[d=\sqrt{16\cos^2(\alpha) + 16 \sin^2(\alpha)}\]

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