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3. The velocity of a skydiver (in feet per second) at a time t seconds after jumping out of an airplane is given by the function v(t) = 176(1 - e^{-0.02t}) (a) Use a definite integral to find the skydiver's average velocity over the time interval [10, 50] seconds. Hint: The answer is between 70 and 80 feet per second. (b) The position is the antiderivative of the velocity. Find the postion function s(t) by integrating v(t), then use the position function to find how far the skydiver has fallen during the first 20 seconds after their jump. Hints: • Choose the value for the constant of integration so that s(0) = 0, then s(t) will measure the distance (in feet) between the airplane and the skydiver t seconds after the dive. • The answer is between 600 and 700 feet. 

          3. The velocity of a skydiver (in feet per second) at a time t seconds after jumping out of an airplane is given by the function

v(t) = 176(1 - e^{-0.02t})

(a) Use a definite integral to find the skydiver's average velocity over the time interval [10, 50] seconds. Hint: The answer is between 70 and 80 feet per second.

(b) The position is the antiderivative of the velocity. Find the postion function s(t) by integrating v(t), then use the position function to find how far the skydiver has fallen during the first 20 seconds after their jump. Hints:
• Choose the value for the constant of integration so that s(0) = 0, then s(t) will measure the distance (in feet) between the airplane and the skydiver t seconds after the dive.
• The answer is between 600 and 700 feet.
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3. The velocity of a skydiver (in feet per second) at a time t seconds after jumping out of an airplane is given by the function v(t) = 176(1 - e^-0.02t) (a) Use a definite integral to find the skydiver's average velocity over the time interval [10, 50] seconds. Hint: The answer is between 70 and 80 feet per second. (b) The position is the antiderivative of the velocity. Find the postion function s(t) by integrating v(t), then use the position function to find how far the skydiver has fallen during the first 20 seconds after their jump. Hints: • Choose the value for the constant of integration so that s(0) = 0, then s(t) will measure the distance (in feet) between the airplane and the skydiver t seconds after the dive. • The answer is between 600 and 700 feet.

Added by Barbara F.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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3. The velocity of a skydiver (in feet per second) at a time t seconds after jumping out of an airplane is given by the function v(t) = 176(1 - e^{-0.02t}) (a) Use a definite integral to find the skydiver's average velocity over the time interval [10, 50] seconds. Hint: The answer is between 70 and 80 feet per second. (b) The position is the antiderivative of the velocity. Find the postion function s(t) by integrating v(t), then use the position function to find how far the skydiver has fallen during the first 20 seconds after their jump. Hints: * Choose the value for the constant of integration so that s(0) = 0, then s(t) will measure the distance (in feet) between the airplane and the skydiver t seconds after the dive. * The answer is between 600 and 700 feet.
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Transcript

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00:01 So for this problem, we are given the velocity of the skydiver.
00:07 So we're going to start by using a definite integral to find their average velocity over the time interval.
00:17 So that's going to be average velocity.
00:21 We're going to use the average value of the function formula.
00:24 So that's 1 over b minus a, which is going to be 50 minus 10 times the integral from a to b to b of the actual function.
00:31 With respect to t.
00:33 So that's going to be 176 times 1 minus e to the negative 0 .02t.
00:40 Dt.
00:41 Then we can evaluate this just using a calculator.
00:47 That's 1 over 40 times the integral from 10 to 50 of 176 times 1 minus e raised to the negative 0 .02 t d t.
01:04 So that's going to give us 76 .8, which we were told, and that's in feet per second, which we're told the answer is between 70 and 80.
01:11 So we got that right.
01:13 Then we want to find the antiderivative of the velocity...
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