Class 1 - Motion in One Dimension Introduction Average Velocity - - PDF document

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 =

5.2km 43km/h + 27min = 0.12h + 0.45h = 0.57h

(c) vx = ∆x

∆t = 6.4km 0.57h = 11km/h 7

Instantaneous Velocity

The instantaneous velocity is the velocity at a particular instant in time. vx = lim

∆t→0

∆x ∆t = dx dt

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Graphical Representation of Instantaneous Velocity

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−40m 40s−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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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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