A roller coaster moves 85m horizontally then travels

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5) A roller coaster moves 200 m horizontally and then travels 135 m at an angle of 30.0 degrees above the horizontal. It next travels 135 m at an angle of 40.0 degrees below the horizontal. What is its displacement from its starting position? (421 m @ -2.63 degrees)

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So we assume that the horizontal is positive x and the displacement is 100 meters horizontally in x axis so the x and y component of x. That would be equal to 100 meter y, which 0, because it's moving in horizontal axis now let b equals to 135 equals at 30 degree horizontal, so now to find a displacement in x direction. That would be 1. He 35 cost 30 degree. This implies 116.9 meters. Similarly, b y o 125 sine of 30 degree that 67.5 meter now exactly like that, we find the third part that would be c c is equal to its given again 135 meat 40 degree. Now that is in the downward direction, we will find c x. U 35 cos of 40 degree. This implies 103.42 c y would be minus of 1 hude 35 problem point indicative y direction into sine of 40 degrees. This implies minus 9 to 88 meters 86.78. So now, let's find the resultant of x comp. Let'S assume our x is equal to x, plus d x x. This implies 100, this 6.9 plus 3.4, that 420.33 resultantiy direction. Wilayat is equal to plus 67.5 plus minus 8678. Is it minus of 19.8 meters, so we have our x and we have r y. Now, let's find the total displacement, total displacement would be the resultant of r x, not y. So let's assume the resultant to be r that is under root of x, squared plus r square. This would give under root of 420.33 square plus minus 90.8 square gives 420 approximately 4 ur 20 minutes. Now, let's find the angle at which is an theta is equal to cori upon it. Therefore, theta is equal to tan. Inverse minus 19.28 upon 420 in 3 implies the equal to minus .363 degrees. So here is the answer: 421 meter and theta equal to minus 2.63.

Problems – 1 A roller coaster moves 85 m horizontally, then travels 45 m at an angle of 30. 0° above the horizontal. What is its displacement from its starting point? Use graphical techniques (i. e. draw a scale diagram). 10° 30° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 2 A bird dog on the trail of a grouse walks 65 m along a heading of 75°, then turns to walk 48 m along a heading of 27°. Use graphical techniques to find the resultant displacement of the dog. Dx 1 = 65 m @ 75 o (15° North of East) Dx 2 = 48 m @ 27 o (27° East of North) Dxr = ? Scale: 1 cm = 10 m Dx 2 (4. 8 cm) = 48 m @ 27° 0 o 270 o Dxr (11 cm) 90 o 0 o 180 o 0 o 35° 90 o Dx 1 (6. 5 cm) = 65 m @ 75° Dxr = 110 m @ 35 o North of East Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 3 A novice pilot sets a plane’s controls, thinking the plane will fly at 250 km/h to the north. If the wind blows at 75 km/h toward the southeast (@135°), what is the plane’s resultant velocity? Use graphical techniques. 45° 15° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 4 While flying over the Grand Canyon, the pilot slows the plane down to one-half the velocity in item 3. If the wind’s velocity is still 75 km/h toward the southeast, what will the plane’s new resultant velocity be? 45° 36° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 5 A dog searching for a bone walks 3. 50 m south, then 8. 20 m at an angle of 30. 0 north of east, and finally 15. 0 m west. Use graphical techniques to find the dog’s resultant displacement vector. 4. 3° 30° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 6 While following the directions on a treasure map, a pirate walks 45. 0 m north and then turns and walks 7. 5 m east. What single straight-line displacement could the pirate have taken to reach the treasure? Use graphical techniques. 10° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 7 While following the directions on a treasure map, a pirate walks 45. 0 m north and then turns and walks 7. 5 m east. What single straight-line displacement could the pirate have taken to reach the treasure? q Copyright © by Holt, Rinehart and Winston. All rights reserved.

Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 9 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 10 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 11 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 12 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 14 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 17 A roller coaster moves 85 m horizontally, then travels 45 m at an angle of 30. 0° above the horizontal. What is its displacement from its starting point? Use algebraic methods. 30° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 18 A novice pilot sets a plane’s controls, thinking the plane will fly at 250 km/h to the north. If the wind blows at 75 km/h toward the southeast (@135°), what is the plane’s resultant velocity? Use algebraic methods. 45° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 19 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. Use algebraic methods to find the magnitude and direction of the runner’s total displacement. 25° Copyright © by Holt, Rinehart and Winston. All rights reserved.

Projectiles Launched Horizontally The initial velocity is the initial horizontal velocity. The initial vertical velocity is 0 (vy, i = 0). Horizontal Component Vertical Component (Constant Velocity) (Constant Acceleration) vx = vx, i = constant Dx = vx · Dt ay = -g = -9. 81 m/s 2 vy, f = ay(Dt) Dy = ½ay(Dt)2 vy, f 2 = 2 ay(Dy) Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 22 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 23 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 24 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 26 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems - 27 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 28 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 29 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 30 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Projectiles Launched at an Angle Resolve the initial velocity into x and y components. The initial horizontal velocity is the xcomponent. The initial vertical velocity is the ycomponent. Horizontal Vertical (Constant Velocity) (Constant Acceleration) vx = vx, i = vicosq = constant Dx = vicosq · Dt ay = -g = -9. 81 m/s 2 vy, f = visinq + ay(Dt) Dy = (visinq)Dt +½ay(Dt)2 vy, f 2 = vi 2(sinq)2 + 2 ay(Dy) Dy = ½(visinq + vy, f )Dt (not included – not necessary) Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 31 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 32 Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 33 vi = ? q Dx = 2. 0 m Dy = 0. 55 m Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 34 The author left this equation out of the list. The problem can be done with the others, but it is much easier with this one. Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 35 1. Yay !! He Makes it!! Copyright © by Holt, Rinehart and Winston. All rights reserved.

Problems – 36 Yeah! He makes it by 4 m-3. 05 m, or 1 m! Copyright © by Holt, Rinehart and Winston. All rights reserved.

Practice Problems 35. Since Vy, f is negative, ball is on its way down. Copyright © by Holt, Rinehart and Winston. All rights reserved.