Physics 218 Alexei Safonov Lecture 3: Kinematics

Motion in One Dimension Where is the car? –X=0 feet at t 0 =0 sec –X=22 feet at t 1 =1 sec –X=44 feet at t 2 =2 sec We say this car has “velocity” or “Speed” Plot position vs. time. How do we get the velocity from the graph?

Motion in One Dimension Cont… Velocity: “Change in position during a certain amount of time” Calculate from the Slope: The “Change in position as a function of time” –Change in Vertical –Change in Horizontal Change:  Velocity   X  t

Constant Velocity Equation of Motion for this example: X = bt Slope is constant Velocity is constant –Easy to calculate –Same everywhere

Moving Car A harder example: X = ct 2 What’s the velocity at t=1 sec? Want to calculate the “Slope” here. Instantaneous Velocity What would my speedometer read?

Displacement Time taken Displacement and Velocity in One Dimension

Definition: Speed = |v(t)| The v(t) vs. t plot is just the slope of the x(t) vs. t plot Displacement and Velocity in One Dimension

Check: Constant Position –X = C = 22 feet –V = slope = dx/dt = 0

Check: Constant Velocity Car is moving –X=0 feet at t 0 =0 sec –X=22 feet at t 1 =1 sec –X=44 feet at t 2 =2 sec What is the equation of motion? X = bt with b=22 ft/sec V = dX/dt  V= b = 22 ft/sec

Derivatives: Common Mistakes The trick is to remember what you are taking the derivative “with respect to” More examples (with a=constant): What if X= 2a 3 t n ? –Why not dx/dt = 3(2a 2 t n )? –Why not dx/dt = 3n(2a 2 t n-1 )? What if X= 2a 3 ? –What is dx/dt? –There are no t’s!!! dx/dt = 0!!! –If X=22 feet, what is the velocity? =0!!!

Clicker Question: Calculate derivative of 2ax+3t 3 x 2 with respect to x? A.2x+3 ● 3 ● t 2 x 2 B.2a+3t 3 ● 2x C.2a+3 ● 3t 2 ● 2x D.2x+3 ● 3t 2 ● 2x

Check: Non-Constant Velocity X = ct 2 with c=11 ft/sec 2 V = dX/dt = 2ct The velocity is: “non-Constant” a “function of time” “Changes with time” –V=0 ft/s at t 0 =0 sec –V=22 ft/s at t 1 =1 sec –V=44 ft/s at t 2 =2 sec

A) YES B) NO Are the plots shown at the left correctly related Displacement and Velocity in One Dimension

The velocity vs. time plot of some object is shown on the right. Which diagram below could be the Displacement vs. time plot for the same object? ABC Clicker Question

Acceleration If the velocity is changing, we are “accelerating” Acceleration is the “Rate of change of velocity” –Just like velocity is the rate of change in distance (or position) You hit the accelerator in your car to speed up –(It’s true you also hit it to stay at constant velocity, but that’s because friction is slowing you down…we’ll get to that later…)

Acceleration Acceleration is the “Rate of change of velocity” Said differently: “How fast is the Velocity changing?” “What is the change in velocity as a function of time?”

Acceleration

Clicker QuestionA B Given displacement and velocity graphs on the left, which graph on the right shows correct dependence of a(t)?

Example You have an equation of motion of: X(t) = X 0 + V 0 t + ½ at 2 where X 0, V 0, and a are constants. What is the velocity and the acceleration?  V(t) = dx/dt = 0 + V 0 + at Remember that the derivative of a constant is Zero!!  Accel = dV/dt =d 2 x/dt 2 = a

Problem with Derivatives A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by –x(t) = bt 2 - ct 3, b= 2.40 m/s 2, c = 0.12 m/s 3 Calculate: –instantaneous velocity of the car at t= 0,5,10 s –How long after starting from rest is the car again at rest?

If the velocity of an object is zero at a given instance, does it mean that the acceleration is zero at the same instance? A.Yes B.No

If the acceleration is zero, does that mean that the velocity is zero? A.Yes B.No

Position, Velocity and Acceleration All three are related –Velocity is the derivative of position with respect to time –Acceleration is the derivative of velocity with respect to time –Acceleration is the second derivative of position with respect to time Calculus is REALLY important Derivatives are something we’ll come back to over and over again

Important Equations of Motion If the acceleration is constant Position, velocity and Acceleration are vectors. More on this in Chap 3

Inverse Derivative? If someone gives us an equation of motion for position x(t), we have shown that by taking derivatives we can find velocity and acceleration Can we go in the opposite order: if someone gives us acceleration, how do we get velocity and position? Is there a reverse of derivative? –Integrals

A car is moving with velocity given by a known formula: v(t)=a+bt+ct 3, with a,b,c being known constants. Is this information sufficient to calculate the position of the car at some specific t=T? A.Yes, position can always be calculated using x(t) = ∫v(t) dt B.No, because one also has to know acceleration (the term in front of t 2 ) C.No, because one also has to know initial position of the object, e.g. at t=0 D.No, because acceleration is not constant

Imagine you were instead given the velocity vs. time graph: –v(t)=v 0 +at You can find the total distance traveled from the area under the curve: –v 0 t + ½at 2 Can also find this by analytical integration… Getting Displacement from Velocity For const acceleration the Equation of motion: x(t)=x 0 +v 0 t + ½at 2

Definite and Indefinite Integrals

At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1 foot between t = 0 sec and t = 1 sec. How far does it move between t = 1 sec and t = 2 sec ? A) 1 foot B) 2 feet C) 3 feet D) 4 feet E) 6 feet Checkpoint 2 Vote again

1ft 4ft 9ft ? 16ft At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1 foot between t = 0 sec and t = 1 sec. How far does it move between t = 1 sec and t = 2 sec ? A) 1 foot B) 2 feet C) 3 feet D) 4 feet E) 6 feet Clicker Question

1ft 4ft 9ft ? 16ft At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1 foot between t = 0 sec and t = 1 sec. How far does it move between t = 1 sec and t = 2 sec ? A) 1 foot B) 2 feet C) 3 feet D) 4 feet E) 6 feet Clicker Question

How quickly can you stop a car? You’re driving along a road at some constant speed, V 0, and slam on the breaks and slow down with constant deceleration a. 1.How much time does it take to stop? 2.How far do you travel before you come to a stop? When you hit the brakes Where you stop

Speeder A speeder passes you (a police officer) sitting by the side of the road and maintains their constant velocity V. You immediately start to move after the speeder from rest with constant acceleration a. How much time does it take to ram the speeder? How far do you have to travel to catch the speeder? What is your final speed? X Police Officer Speeder

Problem Show that for constant acceleration:

Free Fall Free fall is a good example for one dimensional problems Gravity: –Things accelerate towards earth with a constant acceleration –a=g=9.8m/s 2 towards the earth –We’ll come back to Gravity a lot!

Next Time Homework for Chapter 2 due Sunday Pre-lectures for Chapter 3 due Sunday