anonymous
  • anonymous
some calculus anyone? see pic:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
watchmath
  • watchmath
interesting problem. I'll come back later if nobody help you :).
anonymous
  • anonymous
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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amistre64
  • amistre64
the velocity vector of the truck is 4 times the magnitude of the car then right?
amistre64
  • amistre64
add the vectors to get the new vector direction
amistre64
  • amistre64
< 0, y> <4x,4y> -------- <4x, 5y> right?
amistre64
  • amistre64
that should be
amistre64
  • amistre64
new vector is <5x,4y> then
anonymous
  • anonymous
yeah, now im with you...
amistre64
  • amistre64
whenim wring let me know lol
anonymous
  • anonymous
i had to look at the question again :)
amistre64
  • amistre64
the new angle they go off in is y over x
amistre64
  • amistre64
4y ---- = tan(t) , but how to but that in sin cos? 5x
amistre64
  • amistre64
y = sin ; and x = cos right?
anonymous
  • anonymous
yeah so \[4\sin (\alpha)/4\cos(\alpha)= \tan(\theta)\]
amistre64
  • amistre64
truck vector = <4cos(a),4sin(a)> car vecotr = <1,0>
amistre64
  • amistre64
add them to get: < 1 , 0 > <4cos(a),4sin(a)> --------------- <4cos(a) +1, 4sin(a)> = new direction
anonymous
  • anonymous
oh, my numer and denom should be reversed
amistre64
  • amistre64
which is: tan(t) = 4sin(a)/(4cos(a)+1)
amistre64
  • amistre64
b wants you to tell them the velocity, the magnitude, that this new vector is i think
amistre64
  • amistre64
try using distance formula with new vector to see
anonymous
  • anonymous
but tan =cos/sin not tan = sin/cos
amistre64
  • amistre64
scalar the inital velocity of each to u perhaps?
amistre64
  • amistre64
tan = sin/cos ; always has and always will
amistre64
  • amistre64
tan = slope = y/x = sin/cos
anonymous
  • anonymous
dam i looked at cot
anonymous
  • anonymous
bjargh all i can get is \[d=\sqrt{16\cos^2(\alpha) + 16 \sin^2(\alpha)}\]

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