Introduction to Vectors  Introduction to Vectors 2D Kinematics I

Vectors & Scalars A scalar quantity has magnitude but no direction Examples: speed volume temperature mass A vector is a quantity that has both magnitude and direction displacement velocity acceleration force

Vectors in One Dimension With Motion in 1D, our vectors could point in only two possible directions: positive Negative Direction was indicated by the sign (+/-) Examples: +10 m/s meant “to the right” or “up”  9.81 m/s2 meant “to the left” or “down”

Vectors in Two Dimensions   y x In 2D motion, objects can move up-down and left-right north-south and east-west Direction must now be specified as an angle Vectors (drawn as arrows) can represent motion The length of the arrow is proportional to the magnitude of the vector

Vector Addition Vectors can slide around in the plane without changing, but... Changing the magnitude changes the vector Changing the direction changes the vector To add two vectors graphically: Slide them (without changing them) until they are “tip-to-tail” The resultant vector is from the tail of the first to the tip of the second

Adding Perpendicular Vectors To add perpendicular vectors we can: do it graphically or use Pythagorean theorem and the tangent function Let’s add two perpendicular displacements, x and y, to get the resultant displacement d x = 1.5 m right y = 2.0 m down y x x  y tan  = —— x y  = tan–1(——) x y d  = tan–1(———) 1.5 m -2.0 m = -53

Resolving Vectors into Components The vectors x and y are the component vectors of the vector d The x-component vector is always parallel to the x-axis The y-component vector is always parallel to the y-axis x y d

cos  =  =  sin  =  =  x = d cos  y = d sin  We can resolve d back into its components using the cos  and sin  functions: adj hyp x d cos  =  =  adj d y x  opp hyp y d sin  =  =  opp hyp x = d cos  y = d sin  x = d cos  = (2.5 m) cos (-53°) = 1.5 m y = d sin  = (2.5 m) sin (-53°) = -2.0 m

The components of d can be represented by dx and dy dx = x = 1.5 m dy = y = -2.0 m Notice dy is given a negative sign to indicate that y is pointing down

Example of Components Find the components of the velocity of a helicopter traveling 95 km/h at an angle of 35° to the ground

𝒗 𝒙 =𝟗𝟓 𝐜𝐨𝐬 𝟑𝟓° =𝟕𝟕.𝟖𝒌𝒎/𝒉 𝒗 𝒚 =𝟗𝟓 𝐬𝐢𝐧 𝟑𝟓° =𝟓𝟒.𝟓𝒌𝒎/𝒉

Add the following vectors 𝑨 =𝟐.𝟓𝒎/𝒔 𝜽=𝟒𝟓° 𝑩 =𝟓.𝟎𝒎/𝒔 𝜽=𝟐𝟕𝟎° 𝑪 =𝟓.𝟎𝒎/𝒔 𝜽=𝟑𝟑𝟎° Hint: Vectors have both a magnitude and direction. θ is the direction.

𝑨 𝒙 =𝟐.𝟓 𝐜𝐨𝐬 𝟒𝟓°=𝟏.𝟕𝟕𝒎/𝒔 𝑨 𝒚 =𝟐.𝟓 𝐬𝐢𝐧 𝟒𝟓° =𝟏.𝟕𝟕𝒎/𝒔 𝑨 𝒙 =𝟐.𝟓 𝐜𝐨𝐬 𝟒𝟓°=𝟏.𝟕𝟕𝒎/𝒔 𝑨 𝒚 =𝟐.𝟓 𝐬𝐢𝐧 𝟒𝟓° =𝟏.𝟕𝟕𝒎/𝒔 𝑩 𝒙 =𝟓.𝟎 𝒄𝒐𝒔 𝟐𝟕𝟎°=𝟎𝒎/𝒔 𝑩 𝒚 =𝟓.𝟎 𝒔𝒊𝒏 𝟐𝟕𝟎° =−𝟓.𝟎𝟎𝒎/𝒔 𝑪 𝒙 =𝟓.𝟎 𝒄𝒐𝒔 𝟑𝟑𝟎°=𝟒.𝟑𝟑𝒎/𝒔 𝑪 𝒚 =𝟓.𝟎 𝒔𝒊𝒏 𝟑𝟑𝟎° =−𝟐.𝟓𝟎𝒎/𝒔 𝑹 𝒙 = 𝑨 𝒙 + 𝑩 𝒙 + 𝑪 𝒙 =𝟔.𝟏𝟎𝒎/𝒔 𝑹 𝒚 = 𝑨 𝒚 + 𝑩 𝒚 + 𝑪 𝒚 =−𝟓.𝟕𝟑𝒎/𝒔 𝑹= 𝑹 𝒙 𝟐 + 𝑹 𝒚 𝟐 =𝟖.𝟑𝟕𝒎/𝒔 𝜽= tan −𝟏 𝑹 𝒚 𝑹 𝒙 =𝟑𝟏𝟕° 𝐨𝐫 −𝟒𝟑.𝟐°