Question

some calculus anyone?
see pic:

Answer 1

im dying!

Answer 2

interesting problem. I'll come back later if nobody help you :).

Answer 3

maybe i should write my progress too

\[\tan \Theta =((4m \sin \alpha) \div (m + 4 m \cos \alpha)) +1\]

for q a

Answer 4

the velocity vector of the truck is 4 times the magnitude of the car then right?

Answer 5

add the vectors to get the new vector direction

Answer 6

< 0, y>
<4x,4y>
--------
<4x, 5y> right?

Answer 7

that should be <x,0>

Answer 8

new vector is <5x,4y> then

Answer 9

yeah, now im with you...

Answer 10

whenim wring let me know lol

Answer 11

i had to look at the question again :)

Answer 12

the new angle they go off in is y over x

Answer 13

4y
---- = tan(t) , but how to but that in sin cos?
5x

Answer 14

y = sin ; and x = cos right?

Answer 15

yeah so
\[4\sin (\alpha)/4\cos(\alpha)= \tan(\theta)\]

Answer 16

truck vector = <4cos(a),4sin(a)>

car vecotr = <1,0>

Answer 17

add them to get:

< 1 , 0 >
<4cos(a),4sin(a)>
---------------
<4cos(a) +1, 4sin(a)> = new direction

Answer 18

oh, my numer and denom should be reversed

Answer 19

which is: tan(t) = 4sin(a)/(4cos(a)+1)

Answer 20

b wants you to tell them the velocity, the magnitude, that this new vector is i think

Answer 21

try using distance formula with new vector to see

Answer 22

but tan =cos/sin not tan = sin/cos

Answer 23

scalar the inital velocity of each to u perhaps?

Answer 24

tan = sin/cos ; always has and always will

Answer 25

tan = slope = y/x = sin/cos

Answer 26

dam
i looked at cot

Answer 27

bjargh


all i can get is
\[d=\sqrt{16\cos^2(\alpha) + 16 \sin^2(\alpha)}\]