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Relative Motion problem?

  1. Feb 4, 2015 #1
    While traveling on a train, two boys play catch in the aisle. The train is moving north at 30.0 m/s. The ball is tossed front to back at 5.0 m/s relative to the boys. A bystander on the highway observes the ball being tossed toward the back. To the bystander, what is the relative speed of the ball?

    So, I am currently very, very confused in physics. (By the way there's that equation of oVe=oVm+mVe)
    I simply don't understand motion or relative motion or most of the things that come out of my physics teacher's mouth. Okay, so for this answer all I would say is 30-5 just because the ball goes south and the train goes north, so 25? But I don't understand, really, how to put that in terms of the equation or even if it's right. And if it were, what would go with the final answer direction-wise? South, North..? Because obviously the ball is going south...but the train is faster and going north so would it be north as the final answer (from the bystander's point of view?)
     
  2. jcsd
  3. Feb 4, 2015 #2
    Actually you've answered all your questions correctly. You just seem a bit confused. You're right. To the bystander, the ball will appear to travel towards the North witha speed of 25 m/s.

    In terms of an equation, consider the North as positive and South as negative. So the ball will have two velocities: 30 m/s and -5 m/s.
    vf = 30 -5 = 25 m/s
    As the final answer is positive, it means the ball is traveling towards the North
    In case East or West was mentioned, you would have had to do vector addition to find out the final direction.
     
  4. Feb 4, 2015 #3
    To better understand relative motion, consider two cars going side by side at 100 m/s. For a bystander, the two cars will be going at a very high speed. But if the occupants of the two cars look at each other, it will seem that they are stationary. Why? Because the relative motion between them is zero.

    Similarly if you're traveling at say 30 m/s and a car passes you, going in the opposite direction at 30 m/s. It seems like the car is going a lot faster than that. Why? Because you are moving away from that car as well. The relative speed is 30 - (-30) = 60 m/s. The car appears to be going away from you at 60 m/s because of the relative motion between the two.
     
  5. Feb 4, 2015 #4
    First of all, thank you for taking your time to reply. I rather not even ask what vector addition is, or I'll become more confused...but I'm still quite confused. Does "the ball is tossed front to back at 5 m/s relative to the boys" count as the velocity of the ball relative to the train? or relative to the boys? because if it were relative to the train then i could say ballVearth= ballVtrain + trainVearth ...bVe=-5+30. I don't know, I'm just trying to make sense of this.
    Well here's another question: A helicopter is traveling northwest at 210 kph relative to a car. The car is traveling at 30 kph south relative to earth. Find velocity of earth relative to the helicopter. First off, since it says 'northWEST', does that make 210 negative?
     
  6. Feb 4, 2015 #5
    Yes! It is the velocity of the ball relative to the train. The boys are on the train so its one and the same thing.

    You are right

    No! It doesn't make it negative. In this case, you use the formula-
    hve = hvc + cve
    You'll have to resolve the velocity of the helicopter into 2 components. Has the angle of NorthWest been mentioned?
     
  7. Feb 4, 2015 #6

    No, that's all the question said. It ended with "Be careful." actually hahaha. But anyways, I did use that formula and I got hVe= -210+ -30 =-240...and since eVh is the opposite I just made it positive so I got 240 as my final answer. but I'm pretty sure that it's not this simple, especially since the directions are different (northwest and then there's southward)...there's something I'm not doing right here.
     
  8. Feb 4, 2015 #7
    Hehe. No it's not that simple. Have you gone through basic trigonometry? sinΘ and cosΘ?
     
    Last edited: Feb 4, 2015
  9. Feb 4, 2015 #8
    yes I have (I'm in Calculus AB)
     
  10. Feb 4, 2015 #9
    Assume that the angle of the North West speed in 45°. Take the sine and cosine components and then add the components in the same direction. Check the figure I've added.
     

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