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A person skis down a slope with an acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) given by $a=\frac{600 t}{\left(60+0.5 t^{2}\right)^{2}},$ where $t$ is the time (in s). Find the skier's velocity as a function of time if $v=0$ when $t=0$
Calculus 2 / BC
Chapter 26
Applications of Integration
Section 1
Applications of the Indefinite Integral
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Lectures
01:11
In mathematics, integratio…
06:55
In grammar, determiners ar…
02:32
(a) A skier is acceleratin…
04:53
(II) $(a)$ A skier is acce…
04:26
Show that the curve starts…
05:05
03:38
A 60.0-kg skier with an in…
05:14
Skier A 52 -kg skier start…
02:55
00:54
A skier starts from rest a…
06:28
Olympic skier Olympic skie…
02:07
(a) What is the magnitude …
problem. 68. This question is pretty straightforward. There two parts A and B and the scenario that we're given is there is a skier accelerating down a 30 degree hill at 1.80 meters per second squared. And in part of the question, we are asked if I'd be vertical component of her acceleration. So this hill is inclined to 30 degrees and the acceleration of the skier down the hill is 1.80 meters per second squared. So this way down the hill, that straw better. Yeah, and we are asked to find the vertical component of her acceleration, which would be this part right here. Now, looking at this triangle, we can just use basic trigonometry. The sign of 30 degrees is going to be equal to a Y over the high partners, which is 1.8. So to find what a Y is equal to, we just do 1.80 times the sign of 30 degrees. And that gives us a value of a Y equal to about 0.90 zero meters per second squared. So that's part a done now. Part b of the question asked if the elevation changes 335 meters and the skier starts from a position of rest, how much time would it take for her to reach the bottom of the hill? And we have the acceleration in the Y direction that we just calculated of 0.900 meters per second squared. So now that we have this information, we can use one of the kingdom attic equations to solve before time so we can use the equation. We can use this equation. The displacement is equal to the initial velocity times time plus half of acceleration times time squared since the initial lost easier that cancel out to be zero we're left with 335 equals half of accelerations was 0.450 times time squared. So time is just the square root of 335 divided by 0.450 and that gives us an elapsed time. So the time that it takes for this year to reach the bottom of the hill is about 27.3 seconds
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