The velocity of a sky diver is given by the equation v(t) = -54 + 54e^-0.2t Where v is measured in meters per second and t is the amount of time in seconds after the sky diver jumps. Find the change in height of the skydiver in the first 11 seconds as follows 1. Setup the definite integral: ?_a^b Where the bounds of integration are a = b = 2. Find a possible antiderivative for the integrand 3. Use the antiderivative to find the change in height of this sky diver during the first 11 seconds of his jump. (Be accurate to at least one decimal digit.) ?h =
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Setup the definite integral: The change in height is the integral of the velocity function from 0 to 11 seconds. So, we have to evaluate the integral ∫ from 0 to 11 of v(t) dt = ∫ from 0 to 11 of -54e^0.2t dt. Show more…
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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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Madhur L.
Skydiving A skydiver in free fall subject to gravitational acceleration and air resistance has a velocity given by $v(t)=v_{T}\left(\frac{e^{a t}-1}{e^{a t}+1}\right),$ where $v_{T}$ is the terminal velocity and $a>0$ is a physical constant. Find the distance that the skydiver falls after $t$ seconds, which is $d(t)=\int_{0}^{t} v(y) d y$
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