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Chapter 2. Motion Along a Straight Line. Goals for Chapter 2. Describe straight-line motion in terms of velocity and acceleration Distinguish between average and instantaneous velocity and acceleration Interpret graphs position versus time, x(t) versus t (slope = velocity!).
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Chapter 2 Motion Along a Straight Line
Goals for Chapter 2 • Describe straight-line motion in terms of velocityand acceleration • Distinguish between averageand instantaneousvelocity and acceleration • Interpret graphs • position versus time, x(t) versus t • (slope = velocity!)
Goals for Chapter 2 • Describe straight-line motion in terms of velocityand acceleration • Distinguish between averageand instantaneousvelocity and acceleration • Interpret graphs • velocity versus time, v(t) versus t • Slope = acceleration!
Goals for Chapter 2 Understand straight-line motion with constant acceleration Examine freely falling bodies Analyze straight-line motion when the acceleration is not constant
Introduction • Kinematics is the study of motion. • Displacement, velocity and acceleration are important physical quantities. • A bungee jumper speeds up during the first part of his fall and then slows to a halt.
Displacement vs. Distance • Displacement (blue line) = how far the object is from its starting point, regardless of path • Distance traveled (dashed line) is measured along the actual path.
Displacement vs. Distance • Q: You make a round trip to the store 1 mile away. • What distance do you travel? • What is your displacement?
Displacement vs. Distance • Q: You walk 70 meters across the campus, hear a friend call from behind, and walk 30 meters back the way you came to meet her. • What distance do you travel? • What is your displacement?
Displacement vs. Distance • Displacement is written: • SIGN matters! Direction matters! • It is a VECTOR!! Displacement is negative => <= Positive displacement
Speed vs. Velocity Speed is how far an object travels in a given time interval (in any direction) Ex: Go 10 miles to Chabot in 30 minutes Average speed = 10 mi / 0.5 hr = 20 mph
Speed vs. Velocity Velocity includes directional information: VECTOR! Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes Average velocity = 20 mi / 0.333 hr = 60 mph NORTH
Speed vs. Velocity Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes Average velocity = 20 mi / 0.333 hr = 60 mph NORTH
Speed vs. Velocity Velocity includes directional information: VECTOR! Ex: Go 10 miles on 880 Northbound to Chabot in 30 minutes Average velocity = 10 mi / 0.5 hr = 20 mph NORTH
Speed vs. Velocity • Speed is a SCALAR • 60 miles/hour, 88 ft/sec, 27 meters/sec • Velocity is a VECTOR • 60 mph North • 88 ft/sec East • 27 m/s @ azimuth of 173 degrees
Example of Average Velocity • Position of runner as a function of time is plotted as moving along the x axis of a coordinate system. • During a 3.00-s time interval, a runner’s position changes from x1 = 50.0 m to x2 = 30.5 m • What was the runner’s average velocity?
Example of Average Velocity During a 3.00-sec interval, runner’s position changes from x1 = 50.0 m to x2 = 30.5 m What was the runner’s average velocity? Vavg = (30.5 - 50.0) meters/3.00 sec = -6.5 m/s in the x direction. The answer must have value1, units2, & DIRECTION3 Note! Dx = FINAL – INITIAL position
Example of Average SPEED During a 3.00-s time interval, the runner’s position changes from x1 = 50.0 m to x2 = 30.5 m. What was the runner’s average speed? Savg = |30.5-50.0| meters/3.00 sec = 6.5 m/s The answer must have value & units
Negative velocity??? Average x-velocity is negative during a time interval if particle moves in negativex-direction for that time interval.
Displacement, time, and average velocity A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?
Displacement, time, and average velocity—Figure 2.1 Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds? Solution Method: What do you know? What do you need to find? What are the units? What might be a reasonable estimate? DRAW it! Visualize what is happening. Create a coordinate system, label the drawing with everything. Find what you need from what you know
Displacement, time, and average velocity—Figure 2.1 A racing car startsfromrest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds? • “starts from rest” = initial velocity = 0 • car moves along straight (say along an x-axis) • has coordinate x = 0 at t=0 seconds • has coordinate x=+19 meters at t =1 second • Has coordinate x=+277 meters at t = 1+3 = 4 seconds.
Displacement, time, and average velocity—Figure 2.1 Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?
A position-time graph—Figure 2.3 • A position-time graph (an “x-t” graph) shows the particle’s position x as a function of time t. • Average x-velocity is related to the slope of an x-t graph.
Instantaneous Speed Instantaneous speed is the average speed in the limit as the time interval becomes infinitesimally short. Ideally, a speedometer would measure instantaneous speed; in fact, it measures average speed, but over a very short time interval.Note: It doesn’t measure direction!
Instantaneous Speed Instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short. Velocity is a vector; you must include direction!V =27 m/s west…
Instantaneous velocity—Figure 2.4 • The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by vx = dx/dt.
Instantaneous velocity—Figure 2.4 The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by vx = dx/dt. The average speed is not the magnitude of the average velocity!
Instantaneous Velocity Example A jet engine moves along an experimental track (the x axis) as shown. Its position as a function of time is given by the equation x = At2 + B where A = 2.10 m/s2 and B = 2.80 m.
Instantaneous Velocity Example • A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m. • Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s. • Determine the average velocity during this time interval. • Determine the magnitude of the instantaneous velocity at t = 5.00 s.
Instantaneous Velocity Example • A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m. • Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s. • @ t = 3.00 s x1 = 21.7 • @ t = 5.00 s x2 = 55.3 • x2 – x1 = 33.6 meters in +x direction
Instantaneous Velocity Example A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m. b) Determine the average velocity during this time interval. Vavg = 33.6 m/ 2.00 sec = 16.8 m/s in the + x direction
Instantaneous Velocity Example A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m. (c) Determine the magnitude of the instantaneous velocity at t = 5.00 s |v| = dx/dt @ t = 5.00 seconds = 21.0 m/s
Acceleration Acceleration = the rate of change of velocity. Units: meters/sec/sec or m/s^2 or m/s2 or ft/s2 Since velocity is a vector, acceleration is ALSO a vector, so direction is crucial… A = 2.10 m/s2 in the +x direction
Acceleration Example A car accelerates along a straight road from rest to 90 km/h in 5.0 s. What is the magnitude of its average acceleration? KEY WORDS: “straight road” = assume constant acceleration “from rest” = starts at 0 km/h
Acceleration Example A car accelerates along a straight road from rest to 90 km/h in 5.0 s What is the magnitude of its average acceleration? |a| = (90 km/hr – 0 km/hr)/5.0 sec = 18 km/h/sec along road better – convert to more reasonable units 90 km/hr = 90 x 103 m/hr x 1hr/3600 s = 25 m/s So |a| = 5.0 m/s2 (note – magnitude only is requested)
Acceleration Acceleration = the rate of change of velocity.
Acceleration vs. Velocity? • If the velocity of an object is zero, does it mean that the acceleration is zero? • (b) If the acceleration is zero, does it mean that the velocity is zero? Think of some examples.
Acceleration Example An automobile is moving to the right along a straight highway. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration?
Acceleration Example An automobile is moving to the right along a straight highway. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration?
Acceleration Example A semantic difference between negative acceleration and deceleration: “Negative” acceleration is acceleration in the negative direction (defined by coordinate system). “Deceleration” occurs when the acceleration is opposite in direction to the velocity.
Finding velocity on an x-t graph • At any point on an x-t graph, the instantaneous x-velocity is equal to the slope of the tangent to the curve at that point.
Motion diagrams • A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.
Average acceleration • Acceleration describes the rate of change of velocity with time. • The average x-acceleration is aav-x= vx/t.
Instantaneous acceleration • The instantaneous accelerationis ax = dvx/dt. • Follow Example 2.3, which illustrates an accelerating racing car.
Average Acceleration Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average acceleration in the first second? What was its average acceleration in the first 4 seconds?
Findingacceleration on a vx-t graph • As shown in Figure 2.12, the x-t graph may be used to find the instantaneous acceleration and the average acceleration.
A vx-t graph and a motion diagram • Figure 2.13 shows the vx-t graph and the motion diagram for a particle.
An x-t graph and a motion diagram • Figure 2.14 shows the x-t graph and the motion diagram for a particle.
Motion with constant acceleration—Figures 2.15 and 2.17 • For a particle with constant acceleration, the velocity changes at the same rate throughout the motion.
Acceleration given x(t) A particle is moving in a straight line with its position is given by x = (2.10 m/s2)t2 + (2.80 m). Calculate (a) its average acceleration during the interval from t1 = 3.00 s to t2 = 5.00 s, & (b) its instantaneous acceleration as a function of time.