June 16th, 2019, 06:04 PM  #1 
Member Joined: Mar 2019 From: TTF area Posts: 88 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, 07:36 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573 
$v_N = 25\cos(25) = 25\sin(65)$ $v_E = 25\sin(25) = 25\cos(65)$ 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. 
June 16th, 2019, 07:59 PM  #3 
Member Joined: Mar 2019 From: TTF area Posts: 88 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, 08:36 PM  #4 
Newbie Joined: Jun 2019 From: AZ, Seattle, San Diego Posts: 16 Thanks: 10  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, 09:36 PM  #5 
Member Joined: Mar 2019 From: TTF area Posts: 88 Thanks: 1 
That was meant to be but not both.

June 17th, 2019, 06:09 AM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573  Quote:
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)  

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