Physics 207, Lecture 3

Physics 207, Lecture 3 • Today (Finish Ch. 2 & start Ch. 3) • Examine systems with non-zero acceleration (often constant) • Solve 1D problems with zero and constant acceleration (including free-fall and motion on an incline) • Use Cartesian and polar coordinate systems • Perform vector algebra

Acceleration • The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval in which that change occurs. • Bold fonts are vectors • The average acceleration is a vector quantity directed along ∆v

Average Acceleration • Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards. • What was his average acceleration in this time (10 to 20 secs)?

Average Acceleration • Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards. • What was his average acceleration in this time (10 to 20 secs)? N 14 m/s /10 s= 1.4 m/s2 to the SE

Instantaneous Acceleration • Average acceleration • The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zero

x t vx t Position, velocity & acceleration for motion along a line • If the positionx is known as a function of time, then we can find both the instantaneous velocityvxand instantaneous accelerationax as a function of time! ax t

a t 0 Dt tf ti Going the other way…. • Particle motion with constant acceleration • The magnitude of the velocity vector changes Dv = area under curve = a Dt

v1 v0 v3 v5 v2 v4 a v a t 0 Dt t tf ti Going the other way…. • Particle motion with constant acceleration • The magnitude of the velocity vector changes • A particle with smoothly increasing speed: Dv = area under curve = a Dt vf = vi + a Dt = vi + a (tf - ti ) 0

x xi t vx vi t ax t So if constant acceleration we can integrate twice

Two other relationships • If constant acceleration then we also get: Slope of x(t) curve

vx t Example problem • A particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s. • What was the average velocity? • Two legs with constant velocity but …. Slope of x(t) curve

vx t Example problem • A particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s. • What was the average velocity? • Two legs with constant velocity but …. • We must find the displacement (x2 –x0) • And x1 = x0 + v0 (t1-t0) x2 = x1 + v1 (t2-t1) • Displacement is (x2 - x1) + (x1 – x0) = v1 (t2-t1) + v0 (t1-t0) • x2 –x0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6 seconds or 5/3 m/s Slope of x(t) curve

A particle starting at rest & moving along a line with constant acceleration has a displacement whose magnitude is proportional to t2 Displacement with constant acceleration 1. This can be tested2. This is a potentially useful result

Free Fall • When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity. • This acceleration on the Earth’s surface, caused by gravity, is typically written as “little” g • Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i.e. g is a constant.

Gravity facts: • g does not depend on the nature of the material ! • Galileo (1564-1642) figured this out without fancy clocks & rulers! • Feather & penny behave just the same in vacuum • Nominally, g= 9.81 m/s2 • At the equator g = 9.78 m/s2 • At the North pole g = 9.83 m/s2

y When throwing a ball straight up, which of the following is true about its velocity v and its acceleration aat the highest point in its path? Exercise 1Motion in One Dimension • Bothv = 0anda = 0 • v  0, but a = 0 • v = 0, but a  0 • None of the above

When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? Exercise 1Motion in One Dimension • Bothv = 0anda = 0 • v  0, but a = 0 • v = 0, but a  0 • None of the above y

Exercise 2 1D Freefall • Alice and Bill are standing at the top of a cliff of heightH. Both throw a ball with initial speedv0, Alice straightdownand Bill straightup. • The speed of the balls when they hit the ground arevAandvBrespectively. (Neglect air resistance.) • vA < vB • vA= vB • vA > vB Alice v0 Bill v0 H vA vB

Exercise 2 1D Freefall • Alice and Bill are standing at the top of a cliff of heightH. Both throw a ball with initial speedv0, Alice straightdownand Bill straightup. The speed of the balls when they hit the ground arevAandvBrespectively. • vA < vB • vA= vB • vA > vB Alice v0 Bill v0 H vA vB

Problem Solution Method: Five Steps: • Focus the Problem - draw a picture – what are we asking for? • Describe the physics • what physics ideas are applicable • what are the relevant variables known and unknown • Plan the solution • what are the relevant physics equations • Execute the plan • solve in terms of variables • solve in terms of numbers • Evaluate the answer • are the dimensions and units correct? • do the numbers make sense?

A science project • You drop a bus off the Willis Tower (442 m above the side walk). It so happens that Superman flies by at the same instant you release the car. Superman is flying down at 35 m/s. • How fast is the bus going when it catches up to Superman?

yi y 0 t A “science” project • You drop a bus off the Willis Tower (442 m above the side walk). It so happens that Superman flies by at the same instant you release the car. Superman is flying down at 35 m/s. • How fast is the bus going when it catches up to Superman? • Draw a picture

yi y 0 t A “science” project • Draw a picture • Curves intersect at two points

See you Wednesday Assignment: • For Wednesday, Read through Chapter 4.4

Physics 207, Lecture 3