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 June 16th, 2019, 07:04 PM #1 Senior Member   Joined: Mar 2019 From: TTF area Posts: 180 Thanks: 1 Vectors A kite skier is skiing at 25 ms^1 in the N 25° E direction. Calculate the northerly and easterly velocity of the skier. I tried doing Sin 25degrees= Vn/25 Vn= 25 Sin 25degrees= 10.6m/s northernly and doing the same both with Cos I get 22.7m/s EAsternly A strong gust of wind from the west causes the skier to accelerate to the east at 2m/s for 1.8s. CAlculate the new easternly velocity. Use the answer in the first question to calculate the size and direction of the new resultant velocity of the skier. June 16th, 2019, 08:36 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 $v_N = 25\cos(25) = 25\sin(65)$ $v_E = 25\sin(25) = 25\cos(65)$ Updated E component after the acceleration (which is in $m/s^2$ ... pay attention to units) $v_E = 25\sin(25) + (2 \, m/s^2)(1.8 \, sec)$ resultant vector after the acceleration to the East is 26.7 m/s, N 32 deg E. Thanks from helpmeddddd June 16th, 2019, 08:59 PM #3 Senior Member   Joined: Mar 2019 From: TTF area Posts: 180 Thanks: 1 Vn=22.66 ms^1? Northerly Ve=10.57 ms^1? Easterly new Ve= 10.57+3.6= 14.17ms^2 Easternly? what formulas did you use? Thanks. June 16th, 2019, 09:36 PM   #4
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Quote:
 Originally Posted by helpmeddddd … N 25° E direction … … doing the same both with Cos …
Don't use cosine to find both components. We use cosine for the horizontal component (eastward) and sine for the vertical component (northward).

That angle is measured using quadrant bearings, but the trig functions use angles measured in standard position. So, we use the complementary angle (65°).

Vertical component = 25∙sin(65°) ≈ 22.6577

Horizontal component = 25∙cos(65°) ≈ 10.5655 June 16th, 2019, 10:36 PM #5 Senior Member   Joined: Mar 2019 From: TTF area Posts: 180 Thanks: 1 That was meant to be but not both. June 17th, 2019, 07:09 AM   #6
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Quote:
 Originally Posted by helpmeddddd Vn=22.66 ms^1? Northerly Ve=10.57 ms^1? Easterly new Ve= 10.57+3.6= 14.17ms^2 Easternly? what formulas did you use? Thanks.
Make sketches ... Note the velocity's North component is adjacent to the 25 degree angle

$v_N = 25\cos(25) \approx 22.7 \, m/s$

... the velocity's East component is opposite the 25 degree angle

$v_E = 25\sin(25) \approx 10.6 \, m/s$

reference the second sketch for the Easterly acceleration ...

$v_{fE} = v_{0E} + at = 25\sin(25) + (2 \, m/s^2)(1.8 \, sec) \approx 14.2 \, m/s$

$|v_f| = \sqrt{v_{fN}^2 + v_{fE}^2}$

$\theta = \arctan\left(\dfrac{v_{fE}}{v_{fN}}\right)$, where $\theta$ is the angle clockwise from North (complementary to the angle shown in the 2nd sketch) Tags vectors Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post taylor_1989_2012 Physics 2 May 8th, 2016 04:35 AM kuben Geometry 6 April 14th, 2016 04:48 PM Confusion9 Pre-Calculus 4 March 21st, 2015 06:35 AM quantum129 Linear Algebra 3 March 10th, 2015 02:41 PM djiang87 Linear Algebra 8 March 25th, 2010 02:17 AM

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