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1.8 Lesson: Motion in 2D- An Algebraic Approach Addition of Perpendicular Vectors • Two vectors acting perpendicular (900) to each other may be added using algebra instead of a scale diagram. Resultant Displacement of Two Perpendicular Vectors 1. Draw the vector diagram using the rules for vector addition. 2. Use Pythagorean theorem to get the magnitude of the resultant displacement. 3. Use tangent ratio to get the angle of the direction of the resultant vector. Example 1 A car drives 19.0km [S] and then 9.0km [E]. What is the resultant displacement of the car? ∆ = ∆1 + ∆2 ∆ = 19 + 9 [] ∆1 = 19 [] ∆1 ∆ ∆2 ∆1 Applying Pythagoras theorem to calculate the magnitude; ∆ 2 = ∆12 + ∆22 ∆ 2 = 19 2 + 9 2 ∆ 2 = 3612 + 81 2 ∆ = 442 2 ∆ = 21.02 Calculate the angle for the direction ∆2 = ∆2 = 9.0 [] Resultant displacement is 21 km[S 250 E] = ∆ 2 ∆1 9 = 0.474 19 = −1 0.474 = 25.40 Components of a vector • In the previous example you added two perpendicular vectors to obtain one vector called the resultant. • The vector along the horizontal is called the xcomponent of the resultant vector. • The vertical component of the resultant. • A single vector can be broken into components or components combined to form a single vector. Directions Example 1: What are the components of a vector 30.0 m [E 250 N] Set up the directions in the same way as a yaxis versus x-axis graph • E is “+” x-component • W is “-” x-component • N is “+” y-component • S is “-” y-component Component Directions: Example 1 • What are the +[N] components of a vector 30.0 m [E 250 N] −x[W] sin 25° = ∆ ∆ 25° ∆ ∆ cos 25° = ∆ ∆ = ∆ 25° ∆ = 30.0 0.423 ∆ = . [N] +[E] ∆ ∆ ∆ = ∆ 25° ∆ = 30.0 0.906 − [S] ∆ = . [E] Components Method of Vector Addition 1. Draw vectors from the origin of Cartesian coordinate system 2. Break vector into components 2. Solve for total X-component. 3. Solve for total Y-component. 4. Use Pythagorean theorem to get the magnitude of the resultant. 5. Use tangent ratio to get the angle of the direction. Components With Angled Directions: If the x-direction is the reference then: • To get the x-component use cosine ratio • To get the y-component use sine ratio Example 2 A car drives 19.0km[S], 8.0km[E], 6.0km[N], 1.2km[W], and then 2.5km[N]. What is the resultant displacement of the car? A car drives 19.0km [W 30 ̊ N] and then goes 9.0km [N 10 ̊ E] What is the resultant displacement of the car using the components method? +[N] ∆1 ∆1 = ∆1 30° = 19.0 0.5 = 9.5 km ∆2 ∆2 = ∆2 80° = 9.0 0.174 = 1.566 km ∆1 ∆2 ∆1 30° −[W] 10° ∆ 2 80° ∆1 ∆2 −[S] ∆1 = ∆1 30° ∆1 = 19.0 0.866 ∆1 = 16.345 km +[E] ∆2 = ∆2 sin 80 ° ∆2 = 9.0 0.985 = ∆2 = 8.865 km A car drives 19.0km [W 30 ̊N] and then goes 9.0km [N 10 ̊E] What is the resultant displacement of the car using the components method? +[N] ∆ ∆ = ∆1 + ∆2 ∆ = 9.5 + 8.865 ∆ = 18.365 ∆ 2 = ∆ 2 + ∆ 2 ∆ 2 = 14.779 2 + 18.365 ∆ 18.365 km ∆ 2 = 555.69 2 +[E] −[W] ∆ ∆ = 23.57 14.779km ∆ 18.365 −[S] = = ∆ 14. 779 ∆ = ∆1 + ∆2 tan = 1.48 ∆ = −16.345 + 1.566 = −1 1.24 = 51.2° ∆ =− −14.779 Resultant displacement is 23.6 km [W 510 N] 2 Example #4: What is the resultant displacement of an ATV that goes 2.0 km [N], 3.0 km [E], 3.5 km [SE], 5.0 km [20 ̊ N of E], and finally 4.0 km [30 ̊ E from N]. Adding Velocities-River Crossing • A physics student has forgotten her lunch and needs to return home to retrieve it. To do so she hops into her motorboat and steers straight across the river at a constant velocity of 12 km/h [N]. If the river is 0.30 km across and has no current, how long will it take her to cross the river? • Relative Velocity In the second diagram with no current the boat travels directly across the river in a straight line. In the first diagram, the boat’s velocity directly across combined with the river’s current, causes the boat to move diagonally across the river. Adding Velocities-River Crossing • A physics student has forgotten her lunch and needs to • • • • return home to retrieve it. To do so she hops into her motorboat and steers straight across the river at a constant velocity of 12 km/h [N]. The river is 0.30 km across and has current of 24 km/h [E]. Will she reach the shore at her home? What is her resultant velocity across the river? How long will it take her to cross the river? How far from her home will she land? Adding Velocities-River Crossing • = ∆ • ∆ = ∆ • ∆ = 0.30 [] 12 /ℎ[] ∆ • ∆ = 0.025 ℎ 1.5 River Crossing-With Current She will land east of her home. ∆ River Crossing-With Current = 12 /ℎ( ) = 24 /ℎ( ′ ) =? ( ) ∆ 2 = 2 + 2 2 = 12/ℎ 2 + 24/ℎ 2 = 720 /ℎ θ = 720 /ℎ =26.8 km/h 2 2 2 River Crossing-With Current = ∆ θ 24 /ℎ = =2 12 /ℎ = −1 2 = 63° The resultant velocity of the boat is 27 km/h [N 630 E] (2significant figures) River Crossing-With Current To calculate the time to cross the river, focus on the component of the velocity in the direction across the river. This motion is uniform velocity. ∆ = 12 /ℎ[] km [N] ∆ ∆ ∆ = ∆ 0.30 [] ∆ = = 0.025 ℎ 12 [] ℎ Time to cross the river is 0.025 h = θ ∆ =0.30 River Crossing-With Current The boat travels across the river and along the river bank at the same time. To calculate how far from her home the student will land, focus on the motion east. = 24 ∆ = 0.025 ℎ ℎ ∆ θ ∆ =? ∆ = ∆ ∆ = 24 0.025 ℎ ℎ She will land 0.60 km east of her home.