Last time we defined “average velocity” … Then we defined “instantaneous velocity” at a given moment to be the slope of the tangent line at that moment. In general, the velocity changes over time

If it doesn’t, then the position-vs-time graph is just a straight line Not very interesting

How quickly that happens is of great interest to us… It’s more interesting when the instantaneous velocity does change over time. How quickly that happens is of great interest to us…

Just as we defined average velocity as change in position over change in time, we can define average acceleration as the change in velocity over change in time:

Just as we defined average velocity as change in position over change in time, we can define average acceleration as the change in velocity over change in time: Units?

Just as we defined average velocity as change in position over change in time, we can define average acceleration as the change in velocity over change in time: Example 1: An airplane has an average acceleration of 2.19 m/s2 during takeoff. If the airplane starts at rest, how much time does it take to reach 174 mph?

instantaneous acceleration Just as we defined instantaneous velocity to be the slope of the tangent line to the position-vs-time graph, we can also define instantaneous acceleration to be the slope of the tangent line to the velocity-vs-time graph

Graphical interpretation of acceleration An acceleration of 0.25 m/s2 means that the velocity is increasing by 0.25 m/s every second

Graphical interpretation of acceleration An acceleration of - 0.5 m/s2 means that the velocity is decreasing by 0. 5 m/s every second

What does this object do at time t = 2.0 s? A) It stops momentarily, and then continues in the same direction. B) It stops momentarily, and then continues in the opposite direction. C) Nothing. It continues to move in the same direction the whole time.

Example 2: A cyclist riding in a straight line has an initial velocity of 3.5 m/s, and accelerates at – 1.0 m/s2 for 2.0 s. The cyclist then coasts with zero acceleration for 3.0 s, and finally accelerates at 1.5 m/s2 for an additional 2.0 s. What is the final velocity of the cyclist?

Example 2: A cyclist riding in a straight line has an initial velocity of 3.5 m/s, and accelerates at – 1.0 m/s2 for 2.0 s. The cyclist then coasts with zero acceleration for 3.0 s, and finally accelerates at 1.5 m/s2 for an additional 2.0 s. What is the final velocity of the cyclist? What is the average acceleration of the cyclist for these seven seconds?

It is possible to have any combination of signs of velocity and acceleration

It is possible to have any combination of signs of velocity and acceleration

It is possible to have any combination of signs of velocity and acceleration

It is possible to have any combination of signs of velocity and acceleration

Velocity vs. time graphs may help: When the curve is above the t-axis, velocity is positive When v vs t slopes downward, the acceleration is negative Notice the two cases where the velocity may change sign

If the velocity of a car is non-zero (v ≠ 0), can the acceleration of the car be zero? Yes No It depends on the velocity

If the velocity of a car is non-zero (v ≠ 0), can the acceleration of the car be zero? Yes No It depends on the velocity If the acceleration of a car is non-zero (a ≠ 0), can the velocity of the car be zero? Yes No It depends on the velocity

Constant acceleration means the velocity curve is a straight line. Motion with constant acceleration is so important in physics that we will study it specifically Constant acceleration means the velocity curve is a straight line. It also means the average acceleration is the same as the instantaneous acceleration

Let the initial time be 0: This leads to: Which becomes our first “kinematic equation for constant acceleration”

Application: Freefall “Freefall” is when an object moves ONLY under the influence of gravity. The acceleration due to gravity is the same for all objects (if you can ignore sir resistance): g = 9.81 m/s2 It always acts straight downward

Application: Freefall “Freefall” is when an object moves ONLY under the influence of gravity. The acceleration due to gravity is the same for all objects (if you can ignore sir resistance): g = 9.81 m/s2 It always acts straight downward It is common to define upward to be the positive direction (though not necessary) x a = – g

Application: Freefall “Freefall” is when an object moves ONLY under the influence of gravity. The acceleration due to gravity is the same for all objects (if you can ignore sir resistance): g = 9.81 m/s2 It always acts straight downward Example: A rock is thrown straight upward with an initial velocity of 8.6 m/s. What is its velocity after 2.0 s? x a = – g

Let the initial time be zero: To develop another “kinematic equation”, let’s look at the definition of average velocity: Let the initial time be zero:

Let the initial time be zero: To develop another “kinematic equation”, let’s look at the definition of average velocity: Let the initial time be zero: Simple rearrangement leads to:

Remember that we’re talking about constant acceleration here: So our equation becomes

One last adjustment: Remember that v = v0 + at And our final version is

Example: A cyclist is riding at a constant 6.8 m/s when she comes upon a jogger running at about the same speed. In order to pass him, she accelerates at 0.75 m/s2 for 5.6 s. After 5.6 s of constant acceleration: What is her final velocity?

Example: A cyclist is riding at a constant 6.8 m/s when she comes upon a jogger running at about the same speed. In order to pass him, she accelerates at 0.75 m/s2 for 5.6 s. After 5.6 s of constant acceleration: What is her final velocity? How far has she traveled since starting to accelerate?

In that case we can combine these two equations: Sometimes we are not concerned with the elapsed time, just positions, velocities, and acceleration. In that case we can combine these two equations: 1) Solve second equation for t 2) Plug that into the first equation

Sometimes we are not concerned with the elapsed time, just positions, velocities, and acceleration. In that case we can combine these two equations: 1) Solve second equation for t 2) Plug that into the first equation It leads to: Note that this contains no new information!

Example: A car accelerates from 0 to 60.0 mph (26.8 m/s) while covering 75.5 m distance. What is the car’s (assume constant) acceleration?