7
Motion in Two Dimensions

Motion in Two Dimensions

  • Upload
    roch

  • View
    32

  • Download
    0

Embed Size (px)

DESCRIPTION

Motion in Two Dimensions. Example. What is the displacement of a person who walks 10.0 km (E) and then 5.00 km (N) ? D 1 + D 2 = D R Use a “tip to tail” method Draw 1 st vector (D 1 ) Draw 2 nd vector (D 2 ) placing its tail at the tip of the 1 st vector. - PowerPoint PPT Presentation

Citation preview

Page 1: Motion in Two Dimensions

Motion in Two Dimensions

Page 2: Motion in Two Dimensions

Example What is the displacement of a person who

walks 10.0 km (E) and then 5.00 km (N) ? D1 + D2 = DR

Use a “tip to tail” method Draw 1st vector (D1)

Draw 2nd vector (D2) placing its tail at the tip of the 1st vector

Page 3: Motion in Two Dimensions

Draw the arrow from the tail of the 1st vector to the tip of the second. This is the resultant vector (sum of two vectors) - DR

The magnitude of DR is not equal to the sum of D1 and D2

If the two are not in the same direction the resultant is always smaller - D1 + D2 > DR

- DR least if vectors are in opposite directions

- DR most if vectors are in same directions It is not important what order they are added in

Page 4: Motion in Two Dimensions

Method 1 for finding DR

Draw vectors to scale and actually measure the result

10 km

5 kmDR =

Page 5: Motion in Two Dimensions

Method 2 for finding DR

Once the diagram is complete, solve for DR using the Pythagorean theorem

a2 + b2 = c2

10.02 + 5.002 = c2 c = 11.2 km

Page 6: Motion in Two Dimensions

Must also include the direction

Solve for direction of travel using trig identities

Sin θ = opp/hyp cos θ = adj/hyp Sin θ = 5/11.2 θ = 27° Final answer 11.2 km (27º N of E)

Page 7: Motion in Two Dimensions

Example Find the displacement of a car that travels

250 km west, 100 km north, 25 km south, and then 190 km east.