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Displacement, Elapsed Time, Magnitude, Direction, and Average Speed on a Trip.

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    A group of physics students take off in a small plane to visit physics Land. The plan flies at a constant speed of 160 mph. They start at Newtonville and travel 160 miles due north to Faraday Town. At this point they change course for Maxwell City, which is 320 miles away at a heading of 20 degrees North of East. At Maxwell City, they take an hour-long break. They depart for Bohr Village, which is 120 miles away, at a heading of 15 degrees East of South. The final leg takes them 80 miles due south to Einstein City.

    1. What is their displacement for the trip? Give this in terms of how far North/South and East/West they end up from Newtonville.

    2. What was the elapsed time?

    3. What is the magnitude and direction of the average velocity?

    4. Find the average speed for the entire trip.


    2. Relevant equations

    Vx = Vcos(theta)
    Vy = Vsin(theta)

    Average Speed = (total distance)/(total time)

    3. The attempt at a solution

    (First Question)
    So, I think I went a little too far in my calculations and did not quite understand what the problem was asking me. First I figured out all the (x) and (y) components of each of the velocities between each of the cities which is shown below:

    A = From Newtonville to Faraday Town
    B = From Faraday Town to Maxwell City
    C = From Maxwell City to Bohr Village
    D = From Bohr Village to Einstein City

    Vax = 160cos(90) = 0
    Vay = 160sin(90) = 160

    Vbx = 320cos(20) = 300.702
    Vby = 320sin(20) = 109.446

    Vcx = 120cos(255) = -31.06
    Vcy = 120sin(255) = -115.91

    Vdx = 80cos(270) = 0
    Vdy = 80sin(270) = -80

    Then I took all of these individual velocity components and found the overall velocity of the trip by the following:

    V = [(vx)^2 + (vy)^2]^(1/2)

    V = [(0+31.06+300.702+0)^2 + (160+115.91+109.446-80)^2]^(1/2)

    V = 270.94

    Now to find the direction I did the following:

    tan-1(sum of vy/sum of vx) = degrees

    tan-1 (73.536/269.64) = 15.3 degrees

    I concluded that they ended up 270.94 miles away from Newtonville and 15.3 degrees from the East. I got this wrong because I only needed components. Does that mean I should have forgone finding the degrees?

    (Second Question)

    A = Time between Newtonville and Faraday Town
    B = Time between Faraday Town and Maxwell City
    C = Time between Maxwell City and Bohr Village
    D = Time between Bohr Village and Einstein City

    So, First I found the amount of time it took to travel to each city:

    A = 160 miles*(1 hr/160 miles) = 1 hr

    B = 320 miles*(1 hr/160 miles) = 2 hrs

    C = 120 miles*(1 hr/160 miles) = 0.75 hr

    D = 80 miles*(1 hr/160 miles) = 0.5 hr

    Total trip = A + B + C + D

    Total trip = 4.25 hrs.

    Unfortunately I forgot to add the hour long break that they took at Maxwell City, so the total elapsed time should be 5.25 hours. Am I correct? Is there anything else I am missing which could have this problem be incorrect?

    (Question 3)

    So, I thought that since they just wanted the magnitude and direction of the average velocity that I already found this out in question 1. So I said that the magnitude was 270.94 at 15 degrees. But, I was told that I found that 270.94 is the displacement, not the average velocity. What must I do in order to find the average velocity?

    (Question 4)

    So, in order to find the average speed I did the following:

    Average Speed = (Total Distance)/(Total Time)

    Average Speed = (160 miles + 320 miles + 120 miles + 80 miles)/(4.25 hours) = 160 miles/hr

    I was told that I should have included the time it took for the break which is I think the mistake I made for question 2. If I did include the time it took for the break I would get an average speed of 129.524 miles/hour. Is this correct?

    I realize this is a long problem and it really was a challenge for me to calculate. Any assistance would be greatly appreciated. Thank you for your time.
     
  2. jcsd
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