# Calc

The position of an object moving along a line is given by the function s left parenthesis t right parenthesis equals negative 4.9 t squared plus 31 t plus 15. Find the average velocity of the object over the following intervals.

​(a) left bracket 0 comma 6 right bracket
​(b) left bracket 0 comma 5 right bracket
​(c) left bracket 0 comma 4 right bracket
​(d) left bracket 0 comma 1 plus h right bracket​, where hgreater thanminus1 is any real number
​(a) The average velocity of the object over the interval left bracket 0 comma 6 right bracket is ________.

How do you find average velocity given the interval?

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1. average velocity = displacement / time
by the way NOT distance/time
distance is scalar
displacement is vector, might be - part of the time.

if on earth and
original height = 15 meters
original speed up = 31 m/s and v = 31 - 9.8 t
acceleration = -g = -9.8 m/s^2
then
h(t) = 15 + 31 t - (9.8/2)t^2
agreed :)
so here
h(t) = 15 + 31 t -4.9 t^2
1 to 6 seconds
h(6) = 15 +186 - 176.4 =24.6
h(0) = 15
so average v =1.6 m/s
onward - do the rest.
by the way, when does it get to the top?
v = 31 - 9.81 t
v = 0 at top
so t = 3.16 seconds then starts down
If they had asked for SPEED, not Velocity, you would have had to do up from t = 0 to t = 3.16 then down from t = 3.16 to t = 6 and call both the positive and negative distances +

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2. The reason I went into that detail is I am not sure your question is worded correctly. I wonder if they really want the average speed, not the average velocity.

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3. if I go 4 miles from here to downtown at 30 miles/hour, then immediately come back at 30, my average velocity is ZERO. I went nowhere :)
However my average SPEED was 30

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4. Great. Thank you for specifying the details.

I do have another question. How can I find the average velocity over the interval [0, 1+h] given the function -4.9t^2 +31t+15. I am confused how to set this up and how would you do it by the graphing calculator.

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5. at 0 it is 15

at (1+h) it is 15 + 31(1+h) - 4.9(1+h)^2
subtract
d moved = 31(1+h) - 4.9(1+h)^2
divide by time (1+h)
d moved/time = 31 - 4.9(1+h)

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what is average velocity from t = t to t = t+h ?
s(t+h) = 15 + 31 (t+h) -4.9 (t^2+2 t h+h^2)
s(t) = 15 + 31 t - 4.9 t^2
then distance moved is
s(t+h) - s(t) = 31 h - 9.8 t h - 4.9 h^2
divide by time elapsed h
[ s(t+h)-s(t) ]/h = 31 - 9.8 t - 4.9 h

NOW what if I let h ---> 0 ???
That is velocity at t , the definition of the derivative of displacement with time
v = 31 - 9.8 t
(which I automatically used a while ago )
Surely that is what they want you to know.

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