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pathosdebater
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Please Help... A robot travels at a speed of 1.35 m/sec at a direction of 45.0 degrees for 305 seconds. It then travels at a speed of 1.50 m/sec at a direction of 140.0 degrees for 852 seconds. What is the total displacement at the end of the second maneuver?
First, give the y component of the first displacement vector calculated in the problem.
 one year ago
 one year ago
pathosdebater Group Title
Please Help... A robot travels at a speed of 1.35 m/sec at a direction of 45.0 degrees for 305 seconds. It then travels at a speed of 1.50 m/sec at a direction of 140.0 degrees for 852 seconds. What is the total displacement at the end of the second maneuver? First, give the y component of the first displacement vector calculated in the problem.
 one year ago
 one year ago

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Otonashi Group TitleBest ResponseYou've already chosen the best response.0
1308.08m u can use the law of cosines c^2=a^2+b^22ab cos (c) where cos(c)=cos(85degree)
 one year ago

pathosdebater Group TitleBest ResponseYou've already chosen the best response.0
Is 1308.08 the y component or the total displacement?
 one year ago

Otonashi Group TitleBest ResponseYou've already chosen the best response.0
the total
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Firstly, I'm assuming that the degrees are calculated from a absolute degree frame. i.e. North is 000, East is 090. If not, please tell. The first , ycomponent, can be calc'ed by this: \(S_y=u_y t + \frac{1}{2} a_y t^2\) Now, since \(u_y\) will be \(u sin 45^o\)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Then, a_y= 0, t=305, sub it all in to find the \(S_y\) which is the y displacement.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Do you need further help?
 one year ago
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