Chapter One-Dimensional Kinematics
Units of Chapter Position, Distance, and Displacement Average Speed and Velocity Instantaneous Velocity Acceleration Motion with Constant Acceleration Applications of the Equations of Motion Freely Falling Objects
-1 Position, Distance, and Displacement Before describing motion, you must set up a coordinate system define an origin and a positive direction.
-1 Position, Distance, and Displacement The distance is the total length of travel; if you drive from your house to the grocery store and back, you have covered a distance of 8.6 mi.
-1 Position, Distance, and Displacement Displacement is the change in position. If you drive from your house to the grocery store and then to your friend s house, your displacement is.1 mi and the distance you have traveled is 10.7 mi.
Example: A ball is initially at x = + cm and is moved to x = - cm. What is the displacement of the ball? 1 0 1 x (cm) " x = x f! xi =! cm =! 4 cm! cm
Example: At 3 PM a car is located 0 km south of its starting point. One hour later its is 96 km farther south. After two more hours it is 1 km south of the original starting point. (a) What is the displacement of the car between 3 PM and 6 PM? Use a coordinate system where north is positive. x i = 0 km and x f = 1 km " x = x f! xi (! 0 km) = 8 km =! 1 km! +
Example continued (b) What is the displacement of the car from the starting point to the location at 4 pm? x i = 0 km and x f = 96 km " x = x f! xi ( 0 km) = 96 km =! 96 km!! (c) What is the displacement of the car from 4 PM to 6 PM? x i = 96 km and x f = 1 km " x = x f! xi (! 96 km) = 84 km =! 1 km! +
- Average Speed and Velocity The average speed is defined as the distance traveled divided by the time the trip took: Average speed = distance / elapsed time Is the average speed of the red car 40.0 mi/h, more than 40.0 mi/h, or less than 40.0 mi/h?
- Average Speed and Velocity Average velocity = displacement / elapsed time If you return to your starting point, your average velocity is zero.
- Average Speed and Velocity Graphical Interpretation of Average Velocity The same motion, plotted one-dimensionally and as an x-t graph:
-3 Instantaneous Velocity Definition: (-4) This means that we evaluate the average velocity over a shorter and shorter period of time; as that time becomes infinitesimally small, we have the instantaneous velocity.
-3 Instantaneous Velocity This plot shows the average velocity being measured over shorter and shorter intervals. The instantaneous velocity is tangent to the curve.
-3 Instantaneous Velocity Graphical Interpretation of Average and Instantaneous Velocity
Example: Speedometer readings are obtained and graphed as a car comes to a stop along a straight-line path. How far does the car move between t = 0 and t = 16 seconds? Since there is not a reversal of direction, the area between the curve and the time axis will represent the distance traveled.
Example continued: The rectangular portion has an area of Lw = (0 m/s)(4 s) = 80 m. The triangular portion has an area of ½bh = ½(8 s) (0 m/s) = 80 m. Thus, the total area is 160 m. This is the distance traveled by the car.
Average acceleration: -4 Acceleration (-5)
-4 Acceleration Graphical Interpretation of Average and Instantaneous Acceleration:
-4 Acceleration Acceleration (increasing speed) and deceleration (decreasing speed) should not be confused with the directions of velocity and acceleration:
Example: The graph shows speedometer readings as a car comes to a stop. What is the magnitude of the acceleration at t = 7.0 s? The slope of the graph at t = 7.0 sec is a av = " v " t x = v t!! v t 1 1 = ( 0! 0) ( 1! 4) m/s s =.5 m/s
-5 Motion with Constant Acceleration If the acceleration is constant, the velocity changes linearly: Average velocity: (-7)
-5 Motion with Constant Acceleration Average velocity: (-9) Position as a function of time: (-10) (-11) Velocity as a function of position: (-1)
-5 Motion with Constant Acceleration The relationship between position and time follows a characteristic curve.
-5 Motion with Constant Acceleration The three key equations for Ch.:
-6 Applications of the Equations of Motion Hit the Brakes!
Example: A trolley car in New Orleans starts from rest at the St. Charles Street stop and has a constant acceleration of 1.0 m/s for 1.0 seconds. (a) Draw a graph of v x versus t. 16 14 1 10 v (m/sec) 8 6 4 0 0 4 6 8 10 1 14 t (sec)
Example continued: (b) How far has the train traveled at the end of the 1.0 seconds? The area between the curve and the time axis represents the distance traveled. 1! x = v 1 = ( t = 1 sec) "! t ( 14.4 m/s)( 1 s) = 86.4 m (c) What is the speed of the train at the end of the 1.0 s? This can be read directly from the graph, v x = 14.4 m/s.
Example: A train of mass 55,00 kg is traveling along a straight, level track at 6.8 m/s. Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking acceleration has magnitude of 1.5 m/s, can the train be stopped in time? Know: a x = 1.5 m/s, v ix = 6.8 m/s, v fx = 0 Using the given acceleration, compute the distance traveled by the train before it comes to rest. v fx " x = = v ix! v a + ix x a = x " x! = 0 ( 6.8 m/s) (! 1.5 m/s ) = 36 m The train cannot be stopped in time.
-7 Freely Falling Objects Free fall is the motion of an object subject only to the influence of gravity. The acceleration due to gravity is a constant, g.
-7 Freely Falling Objects An object falling in air is subject to air resistance (and therefore is not freely falling).
-7 Freely Falling Objects Free fall from rest:
Example: A penny is dropped from the observation deck of the Empire State Building 369 m above the ground. With what velocity does it strike the ground? Ignore air resistance. y 369 m x a y Given: v iy = 0 m/s, a y = 9.8 m/s, Δy = 369 m Unknown: v yf Use: v fy = v iy + a y! y = a y! y v yf = a y! y
Example continued: v yf = a " y y = (! 9.8 m/s )(! 369) m = 85.0 m/s (downward) How long does it take for the penny to strike the ground? Given: v iy = 0 m/s, a y = 9.8 m/s, Δy = 369 m Unknown: Δt! y! t = = v iy! t +! y a y 1 a! =! y t ay t = 8.7 sec 1
Example: You throw a ball into the air with speed 15.0 m/s; how high does the ball rise? y v iy Given: v iy = +15.0 m/s; a y = 9.8 m/s x a y To calculate the final height, we need to know the time of flight.! y = 1 viy! t + ay! t Time of flight from: v fy = viy + ay! t
Example continued: The ball rises until v fy = 0. v fy " t = v =! iy v a + iy y a y " t = 0 15.0 m/s =!! 9.8 m/s = 1.53 sec The height: 1 " y = viy" t + ay" t = ( )( ) ( 15.0 m/s 1.53s +! 9.8 m/s )( 1.53s) = 11.5 m 1
-7 Freely Falling Objects Trajectory of a projectile:
Summary of Chapter Distance: total length of travel Displacement: change in position Average speed: distance / time Average velocity: displacement / time Instantaneous velocity: average velocity measured over an infinitesimally small time
Summary of Chapter Instantaneous acceleration: average acceleration measured over an infinitesimally small time Average acceleration: change in velocity divided by change in time Deceleration: velocity and acceleration have opposite signs Constant acceleration: equations of motion relate position, velocity, acceleration, and time Freely falling objects: constant acceleration g = 9.81 m/s