2.2 Acceleration

Biblical Reference But flee there quickly, because I cannot do anything until you reach it. Genesis 19:22 Flight of Lot

Bush 2011

Let’s Review VELOCITY is the SLOPE of a distance, position, or displacement vs. time graph.

The RATE of CHANGE of VELOCITY  = Change = FINAL - INITIAL  v = Final velocity – Initial velocity Acceleration – The Definition

Acceleration

The average acceleration of a particle is defined as the change in velocity Dv x divided by the time interval Dt during which that change occurred. v t v1v1 v2v2 t1t1 t2t2 Dv Dt Acceleration is represented by the slope on a velocity-time graph Average Acceleration

(s) x (m) 1234 t t (s) v (m/s) Displacement 25 m Motion Diagrams

The slopes help us sketch the motion of other graphs! Motion Diagrams

t x t v t a Displacement, velocity and acceleration graphs The slope of a velocity-time graph represents acceleration The slope of a displacement-time graph represents velocity Motion Diagrams

t x t v t a Dt Displacement, velocity and acceleration graphs The area under an acceleration-time graph represents change in velocity. Dv The area under a velocity-time graph represents displacement. Dx Motion Diagrams

Comparing and Sketching graphs It is often difficult to use one type of graph to draw another type of graph. t (s) x (m) slope = v List 2 adjectives to describe the SLOPE or VELOCITY The slope is CONSTANT The slope is POSITIVE How could you translate what the SLOPE is doing on the graph ABOVE to the Y axis on the graph to the right? t (s) v (m/s)

Example t (s) x (m) 1 st line 2 nd line 3 rd line The slope is constant The slope is “+” The slope is “-” The slope is “0” t (s) v (m/s)

Example – Graph Matching t (s) v (m/s) t (s) a (m/s 2 ) t (s) a (m/s 2 ) t (s) a (m/s 2 ) What is the SLOPE(a) doing? The slope is increasing

Kinematics - Analyzing motion under the condition of constant acceleration. Δd or Δx or ΔyChange in Displacement tTime V i or V o Initial Velocity V f or VFinal Velocity aAcceleration gAcceleration due to gravity (9.8 m/s 2 )

Equation #EquationMissing Variable Equation 1 Δx Equation 2 VfVf Equation 3 a Equation 4 t Equation 5 ViVi

An object moving with an initial velocity v i undergoes a constant acceleration a for a time t. Find the final velocity. Solution: Eq 1 vivi time = 0time = t a ? 1-D Motion, Constant Acceleration

What are we calculating? 0 t a DV 1-D Motion, Constant Acceleration

An object moving with a velocity v o is passing position x o when it undergoes a constant acceleration a for a time t. Find the object’s displacement. time = 0time = t xoxo ? a vovo Eq 2 1-D Motion, Constant Acceleration Solution:

What are we calculating? 0 t vovo v 1-D Motion, Constant Acceleration

Eq 1 Eq 2 Solve Eq 1 for a and sub into Eq 2: Solve Eq 1 for t and sub into Eq 2: Eq 3 Eq 4 1-D Motion, Constant Acceleration

1. Read the problem carefully. 2. Sketch the problem. 3. Visualize the physical situation. 4. Identify the known and unknown quantities. 6. Use the variable that is not needed to pick the appropriate Big 5 equation. 7. Solve the equation for the variable needed. 9. Solve and check your answers. 5. Determine the variable that is not needed. 8. Plug quantities into equation. Review – Problem Solving Skills

Example: A Cessna Aircraft goes from 0.00 m/s to 60.0 m/s in 13.0 seconds. Calculate the aircraft’s acceleration m/s 2 Given: v i = 0.00 m/s, v f = 60.0 m/s, Dt = 13.0 sec Find: a (acceleration)

Example: The Cessna now decides to land and goes from 60.0 m/s to 0.00 m/s in 11.0 s. Calculate the Cessna’s acceleration. deceleration m/s 2 Given: v i = 60.0 m/s, v f = 0.00 m/s, Dt = 11.0 sec Find: a (acceleration)

Example: You are designing an airport for small planes. One kind of plane that might use this airfield must reach a velocity before take off of at least 27.8 m/s and can accelerate at 2.00 m/s 2. a) If the runway is 150m long, can this airplane reach the required speed for take off? b) If not, what minimum length must the runway have?

Given: v i = 0.00 m/s, v f = 27.8 m/s, a = 2.00 m/s 2 Verify: Dx < 150. m The runway is too short.

Example: A roller coaster designer is designing a new roller coaster. A section of the coaster requires the cars that are traveling at 23.6 m/s must stop in 0.80 m. The roller coaster designer needs to know if the amount of “g” force will be too great on the passengers. A “g” force of 16 g can be deadly g = 9.8 m/s 2 16 g = m/s 2

Given: v i = 23.6 m/s, v f = 0.00 m/s, Dx = 0.80 m Verify: a < m/s 2 The g-force is too great.

Given: a = 156 m/s 2, v f = 0.00 m/s, Dx = 0.80 m Find: v i 15.8 m/s is a safe speed. What is the maximum velocity that can safely stop within 0.80 m?

Franz Reuleaux