practice ‘c’ p. 94 #’s 1, 2 & 3

1. A football player runs directly down the field for 35 m before turning to the right at an angle of 25° from his original direction and running an additional 15 m before getting tackled. What is the magnitude and direction of the runner’s total displacement? 35 m 25o 15 m sin Θ = opposite = y hypotenuse h cos Θ = adjacent = x hypotenuse h y = (h)(sin Θ) = (15 m) (sin25) = (15 m) (.423) = 6.3 m x = (h)(cos Θ) =(15 m) (cos 25) = (15 m) (.906) = 13.6 m

Now find the resultant of a NEW triangle tan Θ = opp adj Pyth theorem: h2 = x2 + y2 Θ = 6.3 m = 0.1296 = 7o 48.6 m h2 = (48.6 m)2 + (6.3 m)2 h2 = 2361 m2 + 40 m2 35 m + 13.6 m = 48.6 m = x h2 = 2401 m2 Θ ?? 25o 6.3 m = y 49 m h = 49 m h = 49 m to the right of downfield at 7o

2. A plane travels 2.5 km at an angle of 35° to the ground and then changes direction and travels 5.2 km at an angle of 22° to the ground. What is the magnitude and direction of the plane’s total displacement? 5.2 km 22o 2.5 km 35o

2. A plane travels 2.5 km at an angle of 35° to the ground and then changes direction and travels 5.2 km at an angle of 22° to the ground. What is the magnitude and direction of the plane’s total displacement? For the 1st “leg” 5.2 km 22o cos Θ = adjacent = x hypotenuse h 2.5 km 35o x = h cos Θ = (2.5 km) (cos 35) = (2.5 km) (.819) = 2 km sin Θ = opposite = y hypotenuse h y = h sin Θ = (2.5 km) (sin 35) = (2.5 km) (.574) = 1.4 km

Need to find resultant magnitude & direction 2. A plane travels 2.5 km at an angle of 35° to the ground and then changes direction and travels 5.2 km at an angle of 22° to the ground. What is the magnitude and direction of the plane’s total displacement? For the 2nd “leg” 5.2 km 22o cos Θ = adjacent = x hypotenuse h 2.4 km 35o x = h cos Θ = (5.2 km) (cos 22) = (5.2 km) (.927) = 4.82 km sin Θ = opposite = y hypotenuse h y = h sin Θ = (5.2 km) (sin 22) = (5.2 km) (.375) = 1.95 km Need to find resultant magnitude & direction

h = 7.6 km = Θ h2 = x2 + y2 h2 = (6.82 km)2 + (3.35 km)2 2. A plane travels 2.5 km at an angle of 35° to the ground and then changes direction and travels 5.2 km at an angle of 22° to the ground. What is the magnitude and direction of the plane’s total displacement? For the final resultant h 5.2 km 22o 2.4 km y = 1.4 km + 1.95 = 3.35 km Θ h2 = x2 + y2 2 km + 4.82 km = x h2 = (6.82 km)2 + (3.35 km)2 6.82 km = x h2 = 47 km2 + 11 km2 h2 = 58 km2 h = 7.6 km =

For Θ Θ = 26o above the horizon h = 7.6 km 2. A plane travels 2.5 km at an angle of 35° to the ground and then changes direction and travels 5.2 km at an angle of 22° to the ground. What is the magnitude and direction of the plane’s total displacement? 5.2 km 22o For Θ 2.4 km y = 3.35 km Θ tan Θ = opp adj 6.78 km = x Θ = tan -1 opp adj Θ = 26o above the horizon Θ = tan -1 3.35 km = .4911 6.82 km h = 7.6 km

3. During a rodeo, a clown runs 8 3. During a rodeo, a clown runs 8.0 m north, turns 55° north of east, and runs 3.5 m. Then, after waiting for the bull to come near, the clown turns due east and runs 5.0 m to exit the arena. What is the clown’s total displacement? 5 m 3.5 m For the 2nd vector: 55o x = h cos Θ = (3.5 m) (cos 55) = (3.5 m) (.574) = 2 m 8 m cos Θ = adjacent = x hypotenuse h y = h sin Θ = (3.5 m) (sin 55) = (3.5 m) (.819) = 2. 9 m sin Θ = opposite = y hypotenuse h

h2 = x2 + y2 h2 = (7 m)2 + (10.9 m)2 h2 = 49 m2 + 121 m2 h2 = 170 m2 3. During a rodeo, a clown runs 8.0 m north, turns 55° north of east, and runs 3.5 m. Then, after waiting for the bull to come near, the clown turns due east and runs 5.0 m to exit the arena. What is the clown’s total displacement? h = 13 m h2 = x2 + y2 2 m + 5 m = 7 m = x h2 = (7 m)2 + (10.9 m)2 55o h2 = 49 m2 + 121 m2 h2 = 170 m2

h2 = x2 + y2 h2 = (7 m)2 + (10.9 m)2 h2 = 49 m2 + 121 m2 h2 = 170 m2 3. During a rodeo, a clown runs 8.0 m north, turns 55° north of east, and runs 3.5 m. Then, after waiting for the bull to come near, the clown turns due east and runs 5.0 m to exit the arena. What is the clown’s total displacement? Use the x & y vectors to find resultant h: 2 m + 5 m = 7 m = x h2 = x2 + y2 55o h2 = (7 m)2 + (10.9 m)2 8m + 2.9 m = 10.9 m = y h2 = 49 m2 + 121 m2 h = 13 m h2 = 170 m2

3. During a rodeo, a clown runs 8 3. During a rodeo, a clown runs 8.0 m north, turns 55° north of east, and runs 3.5 m. Then, after waiting for the bull to come near, the clown turns due east and runs 5.0 m to exit the arena. What is the clown’s total displacement? For Θ use the vectors… Θ = tan -1 opp adj h = 13 m 55o 10.9 m = y = 10.9 m = 1.557 = 57o 7 m Θ 7 m = x Θ = 57o north of east