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Chapter 4: Motion in Two Dimensions ​by Rijul Garg Background Two dimensional motion describes the displacement, velocity, and acceleration of objects in two dimensions, typically along the x-axis and y-axis. Two Main Strategies for Two Dimensional Motion 1. Consider the x and y dimensions separately using kinematic equations. Resolve individual components to find final answer. 2. Use vector-based kinematic equations in unit-vector notation. Formulas to Know v f = v i + at v +v
v = f2 i △x = v i t + 12 at2 v f 2 = v i 2 + 2a△x v = dx
dt Projectile Motion 1. An object thrown in the air is considered a projectile, and its movement can be tracked using kinematics. 2. In the y-direction, there is a force due to gravity which creates a downward acceleration of 9.8 m/s​2​. 3. We usually assume that this acceleration due to gravity is constant and that air friction is negligible. 4. Typically, there is no force accelerating the object in the x-direction. 5. Remember to simplify the problem by considering x and y-motion separately. Lab: AP Review Sheets
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Circular Motion 1. Because acceleration is either a change in speed or change in direction, an object traveling in a circle at a constant speed still experiences a center-seeking acceleration known as the centripetal acceleration. 2. The direction of this acceleration is towards the middle and the magnitude is the velocity of the object squared divided by the radius of the circle (a = v​2​/r). 3. Net acceleration equals radial acceleration added to tangential acceleration (a​net​ = a​rad​ + a​tan​). Relative Motion 1. Relative motion allows us to describe the motion of an object relative to a specific frame of reference. 2. When we are describing relative accelerations, velocities, and positions, we use the following format: velocity of a relative to b plus velocity of b relative to c equals velocity of a relative to c (v​ab​ + v​bc ​= v​ac​). Free Response Problems 1. A piano is thrown from a skyscraper with an initial velocity of 15 m/s at 45° above horizontal. If the building is 50 m tall, how far does the stone travel horizontally? 2. A car of mass 1500 kg drives around a curve with a radius of 25 m. If the road can exert a maximum friction force of 5000 N, how fast can the car travel without skidding? 3. A plane wants to fly at 150 km/h, 135° relative to the ground, but a 48 km/h wind blows from the south. What velocity should the plane fly at relative to the air to reach this desired velocity? Lab: AP Review Sheets
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Solutions Question 1: Δy = v i t + 12 at2 We will start by only considering the y-direction. − 50 = 15sin(45)t + 12 (− 9.8)t2 v iy = v · sin(θ) 0 = − 4.9t2 + 15sin(45)t + 50 Now that we have a quadratic equation we can solve for t. t = 4.752 s Δx = v i t Because the x-acceleration is 0, this is the equation for Δx . Δx = 15cos(45) · 4.752 Plug in and solve! Δx = 37.4 m Question 2: F net = ma We will start with a force analysis. 2
F f riction = m vr The only force accelerating the car centripetally is F​friction​. 2
The acceleration will be m vr , our equation for a​centripetal​. √
v=
√
v=
r · F f riction
m
25 · 5000
1500
Solve for velocity. Plug in to get the answer! This velocity will be the maximum velocity the car can travel at without creating a F​centripetal greater than the F​friction​. v = 9.13 ms Question 3: v plane−ground = v plane−air + v air−ground This equation helps us relate the relative velocities. x : 150cos(135) = v plane−air (x) + 0 Solving in the x-direction first to get v plane−air (x) . v plane−air (x) = − 106 km/h y : 150sin(135) = v plane−air (y) + 48 Solving in the y-direction now to get v plane−air (y) . v plane−air (y) = 56.1 km/h √
√
v =
(− 106)2 + 56.12 = 90.0 km/h Use x2 + y 2 to solve for the velocity. θ = tan−1 ( −106
) = 62.1° We subtract the angle from 180 because our triangle is in the 56.1
second quadrant. 180 - 62.1 = 118°; 90.0 km/h @ 118° Lab: AP Review Sheets
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