Class 1 - Motion in One Dimension

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Class 1 - Motion in One Dimension  Introduction  Average Velocity  Instantaneous Velocity  Acceleration  Homework 1

Transcript of Class 1 - Motion in One Dimension

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Class 1 - Motion in One Dimension

•   Introduction

•  Average Velocity

•  Instantaneous Velocity

•   Acceleration

•  Homework 

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Average Velocity

Consider the motion of the car shown in the figure below.

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Average Velocity (cont’d)

The graph of this motion is shown below.

The average velocity is defined as the distance traveled divided

by elapsed time

vx = ∆x

∆t  =

 xf  − xi

tf  − ti

where ∆x =  xf  −

 xi is called the displacement.

The average velocity is the slope of the line joining the initial

and final points on the position-time graph.

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Example 1: Calculating Average Velocity

From the position versus time graph for the motion of the car,

estimate the average velocity of the car between (a) points A

and B and (b) points C and E.

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Example 1 Solution

From the position versus time graph for the motion of the car,

estimate the average velocity of the car between (a) points A

and B and (b) points C and E.

(a) vx =  ∆x∆t

  =  xf −xi

tf −ti=   55m−30m

10s−0  = 2.5m/s

(b) vx = ∆x

∆t   =

  xf −xi

tf −ti =  −37m−37m

40s−20s   = −

3.7m/s

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Example 2

You drive your BMW down a straight road for 5.2 km at 43

km/h, at which point you run out of gas. You walk 1.2 km

farther, to the nearest gas station, in 27 min. (a) Calculate your

total displacement. (b) Calculate the total elapsed time. (c)

What is your average velocity from the time you started your

car to the time you arrived at the gas station?

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Example 2 Solution

You drive your BMW down a straight road for 5.2 km at 43

km/h, at which point you run out of gas. You walk 1.2 km

farther, to the nearest gas station, in 27 min. (a) Calculate your

total displacement. (b) Calculate the total elapsed time. (c)

What is your average velocity from the time you started your

car to the time you arrived at the gas station?

(a) ∆x = 5.2km + 1.2km  = 6.4km

(b) ∆t =   5.2km43km/h

 + 27min = 0.12h + 0.45h = 0.57h

(c) vx =  ∆x

∆t

  =   6.4km

0.57h

  = 11km/h

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Graphical Representation of InstantaneousVelocity

The instantaneous velocity at a particular instant in time is the

slope of the position versus time graph at that instant.

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Example 3

Estimate the instantaneous velocity of the car at point D in the

position versus time graph below.

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Example 3 Solution

Estimate the instantaneous velocity of the car at point D in the

position versus time graph below.

vx =  ∆x∆t

  =   −40m−40m40s−20s

  = −4.0m/s

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Example 4

The position of a particle moving along the x-axis is given by

x(t) = 7.8 + 9.2t − 2.1t3

with x in meters and t in seconds. What is the velocity of the

particle at t = 3.5 s?

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Example 4 Solution

The position of a particle moving along the x-axis is given by

x(t) = 7.8 + 9.2t − 2.1t3

with x in meters and t in seconds. What is the velocity of the

particle at t = 3.5 s?

vx(t) =   dx(t)dt   = 9.2 − 6.3t2

vx(t = 3.5s) = 9.2 − 6.3(3.5)2 = −68m/s

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Example 5

(a) Your car, starting from rest, gets up to 55 km/h in 3.2 s.

What is its average acceleration?

(b) Later, you brake your car to rest from 55 km/h in 4.7 s.What is its average acceleration in this case?

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Homework Set 1: Due Fri. Sept. 9

•  Read Sections 2.1-2.5

•  Answer Questions 3 & 4

•  Do Problems 1, 4, 9, 13 & 14

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