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pathosdebater

  • 3 years ago

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.

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  1. Otonashi
    • 3 years ago
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    1308.08m u can use the law of cosines c^2=a^2+b^2-2ab cos (c) where cos(c)=cos(85degree)

  2. pathosdebater
    • 3 years ago
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    Is 1308.08 the y component or the total displacement?

  3. Otonashi
    • 3 years ago
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    the total

  4. Shadowys
    • 3 years ago
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    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 , y-component, can be calc'ed by this: \(S_y=u_y t + \frac{1}{2} a_y t^2\) Now, since \(u_y\) will be \(u sin 45^o\)

  5. Shadowys
    • 3 years ago
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    Then, a_y= 0, t=305, sub it all in to find the \(S_y\) which is the y displacement.

  6. Shadowys
    • 3 years ago
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    Do you need further help?

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