Lecture 2. 1D motion with Constant Acceleration. Vertical Motion.

Lecture 2 1D motion with Constant Acceleration. Vertical Motion.

Types of motion Trajectory is the line drawn to track the position of an abject in coordinates space (no time axis). y 1D motion: Trajectory is a straight line. Position can be described by a single position variable, usually x. 2D motion: Trajectory is a curve, but all its points lie in a straight plane. 3D motion: Trajectory is a curve not restricted like the previous two. y 1D motion x 2D motion x

Acceleration = rate of change in velocity Average acceleration: a Dv = D t Instantaneous acceleration: a = lim Δt 0 Unit: m/s 2 Δv Δt The instantaneous acceleration is the slope in v(t) curve.

Acceleration In 1D motion: An object accelerates is it goes faster or slower. In the case of slower, we also say the object decelerates. Acceleration tells us how quickly the velocity changes. how quickly the distance changes.

Acceleration in 1D: slope of v vs t graph In 1D motion: An object accelerates is it goes faster or slower. In the case of slower, we also say the object decelerates. a Dv = > 0 slope positive D t a Dv = < 0 slope negative D t

Constant acceleration in 1D: Formula From Rest: Initial velocity: v 0 = 0 Velocity after a time t : v = at The average velocity : v = v 0 + v = 0 + v 2 2 = at 2 The distance traveled: d = v t = ( at ) 2 t = at 2 2 = 1 2 at 2

Constant acceleration in 1D: Formula From initial velocity not zero: Initial velocity: v 0 0 Velocity after a time t : v = v 0 + at The average velocity : The distance traveled: d = v t = ( v 0 + v v = v 0 + v 2 2 ) t = ( v 0 + (v 0 + at) 2 ) t = ( 2 v 0 2 + at 2 ) t = v 0 t + 1 2 at 2

ACT: Tracking a Train A train car moves along a long straight track. The graph shows the position as a function of time for this train. The graph shows that the train: position 1. speeds up all the time. 2. slows down all the time. time 3. speeds up part of the time and slows down part of the time.

Positive and Negative Acceleration Positive Acceleration = a smile position time Negative Acceleration = a frown position time

Positive vs Negative The car is moving forwards and you step on the gas x > 0 v > 0 a > 0 x x = 0

Positive vs Negative The car is moving forwards and you step on the gas x < 0 v > 0 a > 0 x x = 0

Positive vs Negative The car is moving forwards and you step on the brakes x > 0 v > 0 a < 0 x = 0 x

ACT: Positive vs Negative The car is initially moving backwards and you step on the gas (with gear on forwards). Which of the following is correct? 1. v < 0, a > 0 2. v < 0, a < 0 3. v > 0, a < 0 x = 0 x

Positive vs Negative For 1D motion along a straight line, Speeding up: Velocity and acceleration have same sign. Slowing down: Velocity and acceleration have opposite sign.

Constant acceleration in 1D Formula 2: Velocity vs position, not time! From initial velocity not zero: Initial velocity: v 0 0 Velocity after a time t : v = v 0 + at The average velocity : v = v 0 + v 2 The distance traveled: d = x x 0 = v t = v 0 t+ 1 2 at 2 The final position, x: x = x 0 +v 0 t + 1 2 at 2 What if time is not known? v 2 = v 0 2 +2 a (x x 0 )

Example: Train A train started to move from the rest with the acceleration of 4.0 m/s 2. 1. What is its speed after 6.0 seconds? 2. What is the distance travelled by the train in 6 seconds? 3. At t = 6.0 s, the train started slow down and stopped in 2.0 s. What is the distance traveled by the train during the 2.0 seconds?

Example: Train A train started to move from the rest with the acceleration of 4.0 m/s 2. 1. What is its speed after 6.0 seconds? v = at = (4.0 m/s2 ) (6.0s) = 24m/s 2. What is the distance travelled by the train in 6 seconds? d = at 2 /2 = (½) (4.0m/s 2 )(6.0s) 2 = 72m 3. At t = 6.0 s, the train started slow down and stopped in 2.0 s. What is the distance traveled by the train during the 2.0 seconds? acceleration: a = Δ v 0 24 m/s = = 12 m/s 2 Δ t 2.0 s distance: d = v 0 t + 1 2 a t 2 = (24 m/s) (2.0 s)+ 1 2 ( 12 m/s2 ) (2.0 s) 2 = 24 m

ACT: Acceleration vs Time The acceleration (m/s 2 ) vs. time (s) graph is shown to the right. Draw (velocity vs. time) graph that corresponds to the acceleration graph. Assume velocity initially is 0.

ACT: Acceleration vs Time

Falling objects When objects fall, they increase their speed linearly (constant acceleration). In vaccuum, the falling acceleration is: a = g = 9.8 m/s 2 In vaccuum, the value of acceleration for a falling object does not depend on mass.

Object thrown downward 1. Objects dropped (v 0 = 0) : v = -g t y =y 0 - ½ gt 2 v 2 = -2g (y-y 0 ) + 2. Objects thrown downward (v 0 & v < 0) : v = v 0 - g t y = y 0 +v 0 t - ½ gt 2 - v 2 = (v 0 ) 2-2g (y-y 0 )

Object thrown upward At the highest point: v hp = v 0 + (-g) t But v hp = 0 then: 0 = v 0 - gt time needed to highest point: t = Example: A ball is thrown upward with the velocity of 49 m/s. 1. How long it takes time to reach the highest point? v 0 g 2. How high the ball can reach?

ACT: Ball going up and down When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point of its path? A. Both v = 0 and a = 0 B. v 0 but a = 0 C. v = 0 but a 0

ACT: Throwing an Object Downwards If you drop an object in the absence of air resistance, it accelerates downward at 9.8 m/s 2. If instead you throw it downward, its downward acceleration after release is: A. less than 9.8 m/s 2. B. 9.8 m/s 2. C. more than 9.8 m/s 2.

ACT: Alice and Bob Alice and Bob stand at the top of a cliff of height h. Both throw a ball with initial speed v 0, Alice straight up and Bob straight down. The speed of the balls when they hit the ground are v A and v B, respectively. Which of the following is true? A) v A < v B v 0 v 0 B) v A = v B C) v A > v B v A Alice v B Bob

Most importantly: symmetry! When the Alice s ball passes it initial position, its velocity is v 0 pointing down (just like Bob s) v 0 v 0 v 0 v A Alice v B Bob v A = v B