lecture 05 motion with constant acceleration

Lecture 05: Motion with constant acceleration

Displacement DistanceVelocity SpeedAcceleration

Objectives

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• Familiarize with the four kinematic equations• Solve problems involving motion with

constant acceleration• Define free fall motion• Solve problems involving motion including

free fall motion

Derivation of the kinematic equations

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Derivation of the kinematic equations

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Derivation of the kinematic equations

Kinematic equations for constant acceleration 

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Graphical representation of constant acceleration 

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Guides in problem solving 

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Approach in problem solvingWrite the given and draw the situation.

Determine what was asked.

Determine the right expression to solve the problem.

To check if your answer is right or wrong:See if the value you got makes sense.Check the units.

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Sample Problem:A motorcyclist heading east accelerates after he passes a signpost. His acceleration is constant at 4.0m/s2. At time t = 0 he is 5.0m east of the signpost, moving east at 15m/s.(a) Find his position and speed at time t = 2.0s. (b) Where is the motorcyclist when his velocity is 25m/s?

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To solve for the position x at time t = 2.0s;

Then, for speed at time t = 2.0s:

(a) Find his position and speed at time t = 2.0s.

Given: ax = 4.0m/s2 v0 = 15m/s x0 = 5.0m t = 2.0s

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(b) Where is the motorcyclist when his velocity is 25m/s?

Solving for x and substituting the known values:

Note that time is not given here but we know vx, v0x, ax and x0, therefore we can use:

Evaluate: Do our results make sense?Positive acceleration; velocity is increasing ☺

Given: ax = 4.0m/s2 v0 = 15m/s x0 = 5.0m vx = 25m/s

Example: Young and Freedman, 2.21An antelope moving with constant acceleration covers the

distance between two points 70.0 m apart in 7.00 s. Its speed as it passes the second point is 15.0 m/s.

a) What is its speed at the first point?b) What is its acceleration?

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Hint: Given: x0 – x = 70.0m (assign x0=0) t = 7.0s vx = 15m/s

Equations for motion with constant acceleration:

a) What is its speed at the first point? v0x = ?

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Hint: Given: x0 – x = 70.0m (assign x0=0) t = 7.0s vx = 15m/s

Equations for motion with constant acceleration:

a) What is its speed at the first point? v0x = ?

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Hint: Given: x0 – x = 70.0m (assign x0=0) t = 7.0s vx = 15m/s

Equations for motion with constant acceleration:

a) What is its speed at the first point? v0x = ?

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Hint: Given: x0 – x = 70.0m (assign x0=0) t = 7.0s vx = 15m/s

b) What is its acceleration? ax

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Hint: Given: x0 – x = 70.0m (assign x0=0) t = 7.0s vx = 15m/s

Equations for motion with constant acceleration:

b) What is its acceleration? ax

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Hint: Given: x0 – x = 70.0m (assign x0=0) t = 7.0s vx = 15m/s

Equations for motion with constant acceleration:

b) What is its acceleration? ax

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Hint: Given: x0 – x = 70.0m (assign x0=0) t = 7.0s vx = 15m/s

Solve for ax and substitute the known values:

Example: Young and Freedman, 2.24In the fastest measured tennis serve, the ball left the

racquet at 73.14 m/s. A served tennis ball is typically in contact with the racquet for 30.0 ms and starts from rest. Assume constant acceleration.

a) What was the ball’s acceleration during this serve?b) How far did the ball travel during the serve?

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Hint: Given: vx = 73.14m/s t = 30.0ms = 0.03s v0 = 0 x0 = 0

Equations for motion with constant acceleration:

a) What was the ball’s acceleration during this serve?ax = ?

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Hint: Given: vx = 73.14m/s t = 30.0ms = 0.03s v0 = 0 x0 = 0

Equations for motion with constant acceleration:

a) What was the ball’s acceleration during this serve?ax = ?

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Hint: Given: vx = 73.14m/s t = 30.0ms = 0.03s v0 = 0 x0 = 0

Solve for ax and substitute the known values:

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Equations for motion with constant acceleration:

Hint: Given: vx = 73.14m/s t = 30.0ms = 0.03s v0 = 0 x0 = 0

b) How far did the ball travel during the serve? x = ?

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Equations for motion with constant acceleration:

Hint: Given: vx = 73.14m/s t = 30.0ms = 0.03s v0 = 0 x0 = 0

b) How far did the ball travel during the serve? x = ?

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Hint: Given: vx = 73.14m/s t = 30.0ms = 0.03s v0 = 0 x0 = 0

b) How far did the ball travel during the serve? x = ?

Solve for ax and substitute the known values:

Special case: free fallAll bodies at a particular

location fall with the same downward acceleration regardless of size and weight

Neglect air resistanceDistance of fall is smaller

compared to the radius of the earth

Ignore effects due to the earth’s rotation

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Seatwork: Young and Freedman, Prob. 2.39A flea can jump straight up to a height of 0.440 m, 1) what is the initial speed as it leaves the ground?2) how long is it in the air? (note: when it returned to ground)

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Given: y = 0.440m y0 = 0 g = 9.81m/s2 vy = 0

   Equations for motion with constant acceleration:Equations for motion with constant acceleration:

Seatwork: Young and Freedman, 2.42A brick is dropped (zero initial speed) from the roof of a

building. The brick strikes the ground in 2.50 s. You may ignore air resistance so the brick is in free fall.

3) How tall, in meters, is the building?4) What is the magnitude of the brick’s velocity just before it reaches the ground?

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   Equations for motion with constant acceleration:

Given: t = 2.5s v0y = 0 g = 9.81m/s2 y = 0

Seatwork: Young and Freedman, Prob. 2.39A flea can jump straight up to a height of 0.440 m, 1) what is the initial speed as it leaves the ground?2) how long is it in the air? (when it returned to ground)

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Given: y = 0.440m y0 = 0 g = 9.81m/s2 vy = 0

   Equations for motion with constant acceleration:

1) what is the initial speed as it leaves the ground? v0x = ?

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Given: y = 0.440m y0 = 0 g = 9.81m/s2 vy = 0

   Equations for motion with constant acceleration:

1) what is the initial speed as it leaves the ground? v0y = ?

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Given: y = 0.440m y0 = 0 g = 9.81m/s2 vy = 0

   Equations for motion with constant acceleration:

1) what is the initial speed as it leaves the ground? v0y = ?

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Given: y = 0.440m y0 = 0 g = 9.81m/s2 vy = 0 (at maximum point)

Solve for v0y and substitute the known values:

2) how long is it in the air? t2 = ?

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   Equations for motion with constant acceleration:

 Given: y –y0= 0 g = 9.81m/s2 vy = 0 (at maximum point)

2) how long is it in the air? t2 = ?

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   Equations for motion with constant acceleration:

multiply t*2

use y-y0 = 0

 Given: y –y0= 0 g = 9.81m/s2 vy = 0 (at maximum point)

2) how long is it in the air? t2 = ?

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 Solve for v0y and substitute the known values:

 Given: y –y0= 0 g = 9.81m/s2 vy = 0 (at maximum point)