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Motion vector subtraction, my attempts and where i'm stuck

  1. Aug 21, 2015 #1
    "A cricket ball moving at 4m/s south, is hit over square leg which is due west at 3m/s, what is the change in velocity?"

    The change in velocity ΔV= V-U (where V= final velocity and U= initial velocity)
    The way I was shown is to add negative U to V, which would give us the same triangle and wouldnt change the answer(change in displacement is still 5m) so I believe the answer is just 3m/s-4m/s= -1m/s

    For the second question:
    "A plane is heading due north at 120m/h (Im assuming the teacher made a mistake and it is km/h because 120 m/h is extremely slow)in an easterly crossswind of 50km/h. What direction does the plane have to fly to end up flying due north?"
    My attempt:
    I havent been shown how to do this, but im assuming that the plane needs to fly north east, at an angle greater than what it is now, and i believe we can get the angle now by using SOH CAH TOA:
    tan=opp./adj. (wind at 50km/h is opposite and the 120km/h north is the adjacent) to the angle is tan^-1(50/120) = 22.6 degrees.
    Now to find at what direction the plane must fly to cancel the 50km/h easterly wind, i dont know. Even if I were to add negative 50 to 120 in a graph, the angle would be the same... I would appreciate a hint!

    For the third question:
    "If a car travelling south at 50km/h rounds a bend and travels east at the same speed. What is its change in velocity?"
    ΔV=V-U = 50-50 = 0m/s
    No change in velocity.

    Fourth and last question:
    Just like question 2, I have no idea how to find the angle it should point to. If someone could give me a hint, woudl highly appreciate it :)

     
  2. jcsd
  3. Aug 21, 2015 #2
  4. Aug 21, 2015 #3

    DEvens

    User Avatar
    Education Advisor
    Gold Member

    You have to put homework questions in the template.

    You don't seem to have picked up on the concept of vectors. A vector has a magnitude and a direction. So your third question for example, you have answered that the change in velocity is zero. But that is not the case since the car starts heading south and finishes heading east. The direction has changed. The speed is the same, but the speed is the magnitude.

    Consider a line segment with length 50 cm, extending from the origin at x=0, y=0. That is (0,0). And the other end is at x=50, y=0. That is (50,0). And suppose we make this an arrow by putting an arrow-head on it, at (50,0), and pointing away from the origin. Now rotate this arrow so it points in the +y direction instead of the +x. So the new location of the end is at (0,50) instead of (50,0). What is the change? You must do vector arithmetic, not just subtract the length. The length is still 50. But the vector is now pointing in a different direction. So the change in something that starts at A and finishes at B is B-A. So our vector starts at (50,0), and ends at (0,50). So the change is (0,50) - (50,0) which is just (-50,50).

    So look at your car. Now work out the vector that represents the change in velocity.
     
  5. Aug 21, 2015 #4
    0m/s south 45° east ?
    No woops I dont think I get it. I havent really ever worked with graphs
     
  6. Aug 21, 2015 #5
    Read the link I posted or your text. Attempt a real solution showing the logical steps to your thinking and show us what you've got.
     
  7. Aug 21, 2015 #6
    Okay I will read it after some rest and come back after trying to figure it out :smile:
     
  8. Aug 23, 2015 #7
    I finally understand vectors now. I will do question 2 because it is the one i didnt know how to do and tell me if I'm doing it right now
    "A plane is heading due north at 120km/h in an easterly crossswind of 50km/h. What direction does the plane have to fly to end up flying due north?"
    To find the magnitude in which way the plane is flying, I will square root 120^2 + 50^2 which = 130km/h(is this 130km/h relative to the ground opposite to it?) North (? angle) East
    For the direction to get the angle:
    Since it is a right-angle triangle, I would use tan= opp/adj, 50/120 = 0.4166... tan^-1 (0.4166) = 22.6 degrees
    So the plane is currently going 130km/h North 22.6 degrees East.

    Now to solve the question, I would add -50 to 120, so that I can get the direction it needs to fly to cancel out the 50. The direction would also be 22.6 degrees, but North 22.6 degrees west.
     
    Last edited: Aug 23, 2015
  9. Aug 23, 2015 #8
    Well done.
     
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