Question

The height of an object launched upward from ground level is given by the function s(t) = -4.9t2 + 78.4t, where s(t) is the height in meters and t is the time in seconds. Arrange the time intervals in ascending order of the object's magnitude of average velocity over the interval.

Answer 1

Answer:

• 6.5-8
• 8-11
• 9.5-10
• 9-11
• 4-4.5
• 3-3.5
• 2-2.5

Step-by-step explanation:

I'm always looking for ways to make problems like this simpler than computing 14 function values and doing a bunch of additional arithmetic.

Of course, all that could be done in a spreadsheet, which could then do the sorting for you. However, we'll try some mental math methods.

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The average velocity on an interval for an object in ballistic motion is the slope of the height curve at the midpoint of the interval. The magnitude of velocity is symmetrical about the vertex of the height curve, which occurs at ...

  t = -b/(2a) = -78.4/(2(-4.9)) = 8

That is, the farther the center of the time interval is from t=8, the greater the magnitude of velocity.

Interval: 8-11, 6.5-8, 2-2.5, 4-4.5, 3-3.5, 9-11, 9.5-10

Midpoint: 9.5, 7.25, 2.25, 4.25, 3.25, 10, 9.75

Distance from 8: 1.5, 0.75, 5.75, 3.75, 4.75, 2, 1.75

So, the values of "distance from 8" in order are ...

  0.75, 1.5, 1.75, 2, 3.75, 4.75, 5.75

The corresponding intervals in order of increasing velocity magnitude are ...

• 6.5-8
• 8-11
• 9.5-10
• 9-11
• 4-4.5
• 3-3.5
• 2-2.5

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For your edification, the spreadsheet solution is attached. Having the sort done automatically saved a bunch of time relative to that spent checking and double-checking the work by hand.

The magnitude of the average velocity on the interval [a, b] is

  |v| = |s(b)-s(a)|/(b-a)