Constant velocity Constant acceleration a ms 2 t

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Constant velocity Constant acceleration a, m/s 2 t, s v, m/s 0 t, s

Constant velocity Constant acceleration a, m/s 2 t, s v, m/s 0 t, s x, m t, s

Position vs Time constant acceleration x, m 6 5 4 3 x~t 2 ?

Position vs Time constant acceleration x, m 6 5 4 3 x~t 2 ? 2 1 0 0. 5 1. 0 1. 5 2. 0 2. 5 t, s

Velocity vs Time Displacement? Use the graph to find the object’s displacement from 0

Velocity vs Time Displacement? Use the graph to find the object’s displacement from 0 sec to 2 sec v, m/s 6 5 4 3 Use the graph to find the object’s displacement from 0 sec to 4 sec 2 1 0 1 2 3 4 5 t, s Use the graph to find the object’s displacement from 2 sec to 5 sec

Velocity vs Time Displacement? v, m/s 6 5 4 3 9 meters 2 1

Velocity vs Time Displacement? v, m/s 6 5 4 3 9 meters 2 1 0 1 2 3 4 5 t, s

Velocity vs Time Displacement? v, m/s 6 5 4 3 15 meters 2 1

Velocity vs Time Displacement? v, m/s 6 5 4 3 15 meters 2 1 0 1 2 3 4 5 t, s

Velocity vs Time Displacement? v, m/s 6 5 4 3 13. 75 meters 2

Velocity vs Time Displacement? v, m/s 6 5 4 3 13. 75 meters 2 1 0 1 2 3 4 5 t, s

Velocity vs Time Displacement? v, m/s 6 5 4 3 12. 5 meters 2

Velocity vs Time Displacement? v, m/s 6 5 4 3 12. 5 meters 2 1 0 1 2 3 4 5 t, s

Velocity vs Time v=at v, m/s 5 a ? 0 Displacement? ? 1 2

Velocity vs Time v=at v, m/s 5 a ? 0 Displacement? ? 1 2 3 4 5 t, s

Velocity vs Time v=at v, m/s a. T 5 a 0 a. T 2

Velocity vs Time v=at v, m/s a. T 5 a 0 a. T 2 ½ 25 a 1 2 3 4 T 5 X = ½ at 2 t, s x, m 6 5 4 3 2 1 0 0. 5 1. 0 1. 5 2. 0 2. 5 t, s

v, m/s vf - vi = at acceleration = change in velocity time vf

v, m/s vf - vi = at acceleration = change in velocity time vf X 2 = ½ (vf – vi)T = ½ a. T 2 X 1 = vi. T vi X = X 1 + X 2 = vi. T + ½ a. T 2 0 T t, s Check the units! Xtrain +cases! Xboy Check the limiting a x = vit + ½ at 2 vi = const

v, m/s vf vf - vi = a. T acceleration = change in velocity

v, m/s vf vf - vi = a. T acceleration = change in velocity time X 2 = ½ (vf – vi)T X 1 = vi. T X = X 1 + X 2 =vi T + ½ (vf – vi)T vi 0 T t, s Check the units! Check the limiting cases! X = ½ (vf + vi)T (vf 2– vi 2) X= 2 a 2 a. X = (vf 2– vi 2) T = (vf-vi)/a a≠ 0!

A ball is thrown straight up. At the top of its trajectory, its A.

A ball is thrown straight up. At the top of its trajectory, its A. velocity is zero, acceleration is zero. B. velocity is non-zero, acceleration is non-zero. C. velocity is zero, acceleration is non-zero. D. velocity is non-zero, acceleration is zero.

This graph shows velocity as a function of time for a car of mass

This graph shows velocity as a function of time for a car of mass 1. 5 x 103 kg. What was the acceleration at the 90 s mark? (A) 0. 22 m/s 2 (B) 0. 33 m/s 2 (C) 1. 0 m/s 2 (D) 9. 8 m/s 2 (E) 20 m/s 2

Homework (due Friday) Problems Section 2. 5 1 -D Motion with Constant Acceleration (p.

Homework (due Friday) Problems Section 2. 5 1 -D Motion with Constant Acceleration (p. 50) #30, 39, 40

Quiz A stone is thrown straight upward with an initial velocity v 0. Sketch

Quiz A stone is thrown straight upward with an initial velocity v 0. Sketch graphs: (a) acceleration versus time (b) velocity versus time (c) position versus time.

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