The Physics
Opus in profectus

Miscellaneous Units

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astronomical bodies

light year

  1. Average speed

    This problem is deceptively easy. Averaging is taught in elementary school, which makes this an elementary problem. Right?

    6.0 km/h + 5.0 km/h  = 5.5 km/h
    The wrong method of averaging

    The add-and-divide method of averaging only works when averaging items of equal weight. The average age of the students in a classroom is the sum of their ages divided by the number of students only because each student is considered to have the same significance (a student, is a student, is a student,… ). In this problem, however, the two segments of the walk are significantly different. The second "half" was actually the majority of the walk. It carries more weight than the shorter first "half". Thus, the add-and-divide method won't work.

    Let's return to our definition. Since speed is the rate of change of distance with time, we'll need both the distance traveled and the time it took to complete the walk. After we determine both of these numbers, the rest is easy.

    Δt = 
    Δt1 =  6.0 km  = 1.0 h
    6.0 km/h
    Δt2 =  10 km  = 2.0 h
    5.0 km/h
    v = 
    v =  6.0 km + 10 km  
    1.0 h + 2.0 h  
    v = 5.3 km/h  

    Look closely at the calculations on the right side. Notice that the formula contains Δ (delta) symbols and yet I added the distances in the numerator and the times in the denominator. That's because Δ doesn't mean difference, it means change. During the walk my position didn't change from 6.0 km to 10 km, it changed first by 6.0 km and then by 10 km for a total change of 16 km.

  2. Average velocity

    Velocity is the rate of change of displacement with respect to time. Velocity is a vector, which means the problem should be solved graphically. Draw an arrow pointing toward the top of the page (north). Label it 6 km. Draw another arrow to the left (west) starting from the previous one (arranged head to tail). Make it slightly longer and label it 10 km. Draw a third arrow starting on the tail of the first and ending on the head of the second. Since north and west are at right angles to one another, the resultant displacement is the hypotenuse of a right triangle. Use Pythagorean theorem to find its magnitude and tangent to find its direction.

    Simple geometric drawing

    r =  √[(6.0 km)2 + (10 km)2]  
    r =  11.6619… km  
    tan θ =  opposite  =  10 km
    adjacent 6.0 km
      θ = 59° on the west side of north  

    Divide displacement by time to get velocity.

    v = 
    v = 
    11.6619… km at 59° W of N
    3.0 h
    v = 3.9 km/h at 59° W of N  


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