Question

Each of the following vectors is given in terms of its x- and y-components $v_x = 20$ m/s, $v_y = 41$ m/s. Find the vector's magnitude. Express your answer in meters per second. $v =$ Part B $v_x = 20$ m/s, $v_y = 41$ m/s. Find the vector's direction Express your answer in degrees. $\theta =$ Part C 

          Each of the following vectors is given in terms of its x- and y-components
$v_x = 20$ m/s, $v_y = 41$ m/s. Find the vector's magnitude.
Express your answer in meters per second.
$v =$
Part B
$v_x = 20$ m/s, $v_y = 41$ m/s. Find the vector's direction
Express your answer in degrees.
$\theta =$
Part C
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Each of the following vectors is given in terms of its x- and y-components vx = 20 m/s, vy = 41 m/s. Find the vector's magnitude. Express your answer in meters per second. v = Part B vx = 20 m/s, vy = 41 m/s. Find the vector's direction Express your answer in degrees. θ = Part C

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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v_x = 20 m/s, v_y = 41 m/s. Find the vector's magnitude. Each of the following vectors is given in terms of its x- and y-components. Express your answer in meters per second. Part B v_x = 20 m/s, v_y = 41 m/s. Find the vector's direction. Express your answer in degrees. θ = degrees above +x-axis Part C V = 20 m/s, Vy = 41 m/s. Find the vector's magnitude. Express your answer in meters per second. Each of the following vectors is given in terms of its x- and y-components.
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Transcript

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00:01 Hello, we have to solve the following problem.
00:03 Let's start from part a.
00:05 Here, velocity vector has two components, vx of 20 meters per second and vy of 40 meters per second.
00:15 First, we have to find in part a the magnitude of this factor.
00:21 Let's do this.
00:23 This magnitude is square root of vx squared plus vy squared.
00:59 That is 44 .7 meters per second.
01:06 In part b we have to find the direction.
01:13 Let's do this.
01:18 First of all we will sketch the x and y coordinates because we have to find the direction relative to the x -axis and here both components are positive.
01:31 Therefore the single alpha is a tangent of 40 over 20 which is 63 .4 degree.
01:53 Now let's move on to question c.
02:02 Actually, let's...
02:05 In equation b, we have to re -label this angle as theta instead of alpha.
02:11 And now that is solved.
02:13 So, now let's get back to question c.
02:16 In question c, acceleration has two components, 2 meters per square second and negative 6 mb half meters per square second.
02:31 And first, we have to find the magnitude...
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