Question

. Trained shepherding dogs take the following path to gather the sheep in the field and bring them in for sheering. Let’s consider the lead dog’s motion. First he runs to the edge of the pasture (120, -200)m the he starts working the sheep. 90m at an angle of 160º, then 50m at an angle of 170º, then finally 165m at an angle of 60º. What is the dogs distance of running and what is the displacement?

Answer 1

@IrishBoy123
can I send the drawing on your email, because its not posting here

Answer 2

screen grab it using gyazo

then post a link

its much easier

Answer 3

try this for size


displacement in x will be \(120 + 90 \cos (160) + 50 \cos(170) + 165 \cos(60)\)

in y, \(-200 + 90 \sin (160) + 50 \sin(170) + 165 \sin(60)\)

Answer 4

Displacement
Convert all to rectangular
(120, -200) (120,-200)

(90m, 160*)
90(sin70* more than north 90*) = 84.6 left
90(cos70*) = 30.8 up (-84.6,30.8)
(50m, 170*)
50(sin80* more than north 90*) = 49.2 left
50(cos80* ) = 8.7 up (-49.2,8.7)
(165, 60*)
165(sin60*) = 142.9 up
165(cos60*) = 82.5 right (82.5,142.9)
Add them up…….
(68.7, -17.6)
The question asks for the displacement of running – that would be the hypoteneuse of these rectangular coordinates…….sqrt(68.7^2 + 17.6^2) = 70.9m


Also asked for distance and that would be all the lengths added up
233.23 (the hyp of (120,-200)
90
50
165
Total distance = 538.23

Answer 5

Is this right?

Answer 6

looks good. displacement numbers agree

haven't checked the actual distance number but that is the easy one and method looks good

Answer 7

alright thanks