Chapter 2 – MOTION IN ONE DIMENSION

Chapter 2 – Motion in One Dimension Section 1 – Displacement and Velocity Pages

One Dimensional Motion To simplify the concept of motion, we will first consider motion that takes place in one direction. One example is the motion of a commuter train on a straight track. To measure motion, you must choose a frame of reference. A frame of reference is a system for specifying the precise location of objects in space and time.

One Dimensional Motion For motion equations we will use the following symbols: Dx = horizontal displacement (distance – d) Dy = vertical displacement vi = initial velocity vf = final velocity Dt = time interval a = acceleration

Dx = xf – xi displacement = final position – initial position Displacement is a change in position. Displacement is not always equal to the distance traveled. The SI unit of displacement is the meter, m. Dx = xf – xi displacement = final position – initial position

Average Velocity Average velocity is the total displacement divided by the time interval during which the displacement occurred. In SI, the unit of velocity is meters per second, abbreviated as m/s.

Velocity and Speed Velocity describes motion with both a direction and a numerical value (a magnitude). Speed has no direction, only magnitude. Average speed is equal to the total distance traveled divided by the time interval.

Interpreting Velocity Graphically For any position-time graph, we can determine the average velocity by drawing a straight line between any two points on the graph. If the velocity is constant, the graph of position versus time is a straight line. The slope indicates the velocity. Object 1: positive slope = positive velocity Object 2: zero slope= zero velocity Object 3: negative slope = negative velocity

Passing Lane – DT Graphs

Interpreting Velocity Graphically The instantaneous velocity is the velocity of an object at some instant or at a specific point in the object’s path. The instantaneous velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.

HANDOUT Guess Format!!!!

Quick Check – make sure you are adding formulas to your BLUE card!!! You and your friend are walking to Piston Petes at an average velocity of 0.98 m/s. If it takes you 34 minutes to arrive, what was your displacement?

Quick Check Two students walk in the same direction along a straight path, at a constant speed – one at 0.90 m/s and the other at 1.90 m/s. Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away?

Chapter 2 – Motion in One Dimension Section 2 – Acceleration

Changes in Velocity Acceleration is the rate at which velocity changes over time. An object accelerates if its speed, direction, or both change. Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.

Quick Check – make sure you are adding formulas to your BLUE card!!! With an average acceleration of -1.2 m/s2, how long will it take a cyclist to bring a bicycle with an initial speed of 6.5 m/s to a complete stop?

Sample Problem Given: v = 10.m/s Δt = 4.8 sec a = ? Formula It takes 4.8 seconds for a car’s speed to increase by 10. m/s. What is its acceleration? Given: v = 10.m/s Δt = 4.8 sec a = ? Formula Substitution Answer: a = 2.1 m/s2

Motion with Constant Acceleration When velocity changes by the same amount during each time interval, acceleration is constant. Since velocity, displacement, and acceleration are all vector quantities, you must keep direction in mind when you substitute values into the equations. Up or to the right are considered to be positive directions. Down or to the left are considered to be negative directions.

Hot Wheels Track

Equations for Constantly Accelerated Straight-Line Motion

Quick Check – make sure you are adding formulas to your BLUE card!!! A car is traveling at 50.0 m/s must slow down to 30.0 m/s in the next 10.0 m. What deceleration must the car have? Given: vi = 50.0 m/s vf = 30.0 m/s d = 10.0 m a = ? Formula Substitution Answer:

Sample Problem Answer: t = 0.625 sec A rocket is capable of accelerating at 800. m/s2. How long after lift off will the rocket reach 500. m/s? Given: a = 800. m/s2 vf = 500. m/s vi = 0 m/s t = ? Formula Substitution Answer: t = 0.625 sec

Sample Problem A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

Sample Problem, continued Given: vi = 0 m/s a = 0.500 m/s2 Dx = 4.75 m Unknown: vf = ? Equation: Substitution: Solution: