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 dividedby elapsed time

vx =∆x

∆t=

xf − xi

tf − tiwhere ∆x = xf − xi is called the displacement.

The average velocity is the slope of the line joining the initialand 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 Aand 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 Aand 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 43km/h, at which point you run out of gas. You walk 1.2 kmfarther, to the nearest gas station, in 27 min. (a) Calculate yourtotal displacement. (b) Calculate the total elapsed time. (c)What is your average velocity from the time you started yourcar 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 43km/h, at which point you run out of gas. You walk 1.2 kmfarther, to the nearest gas station, in 27 min. (a) Calculate yourtotal displacement. (b) Calculate the total elapsed time. (c)What is your average velocity from the time you started yourcar 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

= 6.4km0.57h

= 11km/h

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Instantaneous Velocity

The instantaneous velocity is the velocity at a particular instantin time.

vx = lim∆t→0

∆x

∆t=

dx

dt

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Graphical Representation of InstantaneousVelocity

The instantaneous velocity at a particular instant in time is theslope 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 theposition versus time graph below.

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Example 3 Solution

Estimate the instantaneous velocity of the car at point D in theposition 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 theparticle 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 theparticle 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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Acceleration

• Average acceleration

ax =∆vx

∆t=

vxf − vxi

tf − ti

• Instantaneous acceleration

ax = lim∆t→0

∆vx

∆t=

dvx

dt

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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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Example 5 Solution

(a) Your car, starting from rest, gets up to 55 km/h in 3.2 s.What is its average acceleration?

ax =∆vx

∆t=

55km/h − 0

3.2s= 17km/h · s

ax = 17km/h · s

1000m

1km

1h

3600s

= 4.7m/s2

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

ax =∆vx

∆t=

0 − 55km/h

4.7s= −12km/h · s = −3.3m/s2

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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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