Recent content by PleaseAnswerOnegai

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    Understanding a Velocity-Time Graph

    Thank you very much! This helps me greatly. And I have taken Calculus, that would make sense. Thank you again and have a nice day :) Oh, and I wouldn't want to get complacent hahaha, upon calculating: the total displacement in 9s would be -24m, correct?
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    Understanding a Velocity-Time Graph

    And at 2s, the displacement would be 0m because of 2m + (-2m)?
  3. P

    Understanding a Velocity-Time Graph

    I'm really sorry, I meant 2m. Would that be correct?
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    Understanding a Velocity-Time Graph

    Would the displacement at 1s be 2 cm?
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    Understanding a Velocity-Time Graph

    In solving of the triangle, should my solution be 1/2(1)(-4)?
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    Understanding a Velocity-Time Graph

    Based on my understanding: Displacement is a scalar measurement that quantifies how far we have gone from an original point/place/position relative to direction as opposed to Distance which simply measures, well, total "distance." Similarly, velocity takes into account direction. And its...
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    Understanding a Velocity-Time Graph

    Summary:: I think we are still in the earlier parts of Physics and I am confused at how "values" work for a velocity-time graph. We are using the formulas to solve an area of a triangle and rectangle to find the total displacement. If a diagonal line begins from above and continue to go down...
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    Finding the components of this velocity vector

    So it would then be on the 2nd Quadrant, correct? What confused me was the usage of "positive x-axis" so I wrongfully assumed that it would be strictly in the 1st Quadrant (where x values are positive). I will solve it and send my answer! :)
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    Finding the components of this velocity vector

    Hi! What would the correct sketch look like? While I was sketching it I knew something was fishy and I just had to ask hahaha
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    Finding the components of this velocity vector

    Just started Physics and I wasn't sure if my answer and solution was right. I attempted to project it on the cartesian plane based on my understanding. It said "an angle of 160 with the positive x-axis." What I understood from this is that angles of the triangle that makes contact to the x-axis...
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