1. Lesson 2 – Uniform Acceleration

2. Minds-On
The following video shows the relationship between displacement, velocity, and acceleration of a ball tossed into the air

3. Acceleration
Acceleration is defined as the change in velocity over time
Acceleration is caused by a net force acting on an object
Two types of accelerations: instantaneous and average

4. Acceleration
When a person experiences constant acceleration, they can feel it (e.g. a person feels acceleration in a car when it moves from a stop at a light)
People do not feel constant velocity (e.g. people in a car cruising on the highway at 100 km/h in a straight line and constant grade will feel as if they were not moving at all)

5. Instantaneous Acceleration
Instantaneous acceleration is defined as velocity at a specific moment of time
𝑎 = 𝑣 𝑡
In calculus notation:
𝑎 = lim t→0 𝑣 𝛥𝑡

6. Average Acceleration
Average acceleration is defined as the change in velocity over an interval of time
𝑎 𝑎𝑣𝑔 = 𝛥 𝑣 𝛥𝑡 = 𝑣 𝑓 − 𝑣 𝑖 𝑡 𝑓 − 𝑡 𝑖

7. Uniform Acceleration
The type of acceleration we have modelled with the previous equations is uniform acceleration. This means the acceleration is not changing
Question: Describe how you can decide whether a car is moving with uniform acceleration without doing calculations. You have a stopwatch and a police radar gun.

8. Solving Uniform Acceleration Problems
To solve uniform acceleration problems, there are 5 equations (aka The Big 5) which we can use

9. Problem
Consider the following series of events: (1) The cart is at rest against your hand and you begin to exert a steady force on the cart up the incline. (2) The cart leaves contact with your hand as it rolls up the incline. (3) It reaches its highest point. Describe how you can use the chart in the previous slide to help choose the best equation to solve a problem where v 3 = 0 m/s, a 23 = 1.6 m/s2, Δ d 23 = 0.8 m, and Δt23 is the unknown.

10. Describe What’s Happening
What’s happening between (1) and (2) ?
Hand exerts a force on cart
At (1), it accelerates from rest
It reaches a certain velocity at (2)

11. Describe What’s Happening
What’s happening between (2) and (3) ?
At (2), hand ceases to exert a force on cart
It starts decelerating
It reaches a final velocity of 0 m/s at (3)

12. Solution to Problem What variables do we have?
Variable represents
Value
𝑣 3
Final velocity at (3)
0 m/s
𝑎 23
Acceleration from (2) to (3)
1.6 m/s2
Δ 𝑑 23
Displacement from (2) to (3)
0.8 m
What variable are we missing?
Δt23

13. Which Equation to Use?
So which equation can we use?

14. Rearranging the Equation
So, we can use Δ 𝑑 = 𝑣 𝑓 Δ𝑡− 1 2 𝑎 Δ𝑡 2 But we should rearrange the equation to solve for Δ𝑡 1 2 𝑎 Δ𝑡 2 − 𝑣 𝑓 Δ𝑡+Δ 𝑑 =0

15. Solve the Equation
We can now solve the equation using the quadratic formula: Δ𝑡= 𝑣 𝑓 ± 𝑣 𝑓 2 −4 1 2 𝑎 Δ 𝑑 𝑎 Δ𝑡= 0 m/s± (0 m/s) 2 −4 1 2 (1.6 m s 2 ) 0.8m (1.6 m s 2 )

16. Solution
Solving the equation yields: Δ𝑡=±1 s Note that there are 2 solutions: +1 s and -1 s. However, since time positive, we will disregard our -1s result. The final solution is Δ𝑡=1 s

17. Final Statement
To complete the solution, we must describe the solution in writing. Solution: By identifying which variables we are given and which variable we are solving, we concluded that the equation Δ 𝑑 = 𝑣 𝑓 Δ𝑡− 1 2 𝑎 Δ𝑡 2 will help us yield our solution. Solving for Δ𝑡 , the time interval it took for the cart to come to a rest from (2) to (3) is 1s.

18. Exit Ticket
What is the difference between instantaneous acceleration and average acceleration?
What does uniform acceleration mean?
Do people feel constant velocity? Do they feel constant acceleration?