Lesson plan

Topicm9d93858d85c2e49e_1528449000663_0Topic

Non‑uniform rectilinear motion. AccelerationaccelerationAcceleration. Average velocity and instantaneous velocity

Levelm9d93858d85c2e49e_1528449084556_0Level

Second

Core curriculumm9d93858d85c2e49e_1528449076687_0Core curriculum

I. The use of physical concepts and quantities to describe phenomena and to indicate their examples in the surrounding reality.

II. 8) The student uses the concept of acceleration to describe the uniformly accelerated and uniformly decelerated rectilinear motion; determines the acceleration value with the unit; applies to the calculation the relationship between acceleration and the change of speed and time in which this change occurred (Δv = a · Δt).

Timingm9d93858d85c2e49e_1528449068082_0Timing

45 minutes

General learning objectivesm9d93858d85c2e49e_1528449523725_0General learning objectives

Definition of the concept of acceleration and non‑uniform rectilinear motion.

Key competencesm9d93858d85c2e49e_1528449552113_0Key competences

1. Defining the concept of accelerationaccelerationacceleration and deceleration as a vector quantity.

2. Defining the concept of non‑uniform motionnon‑uniform motionnon‑uniform motion.

3. Determination of the unit of acceleration.

Operational (detailed) goalsm9d93858d85c2e49e_1528450430307_0Operational (detailed) goals

The student:

- defines the concept of acceleration and deceleration,

- uses the concept and unit of acceleration in practice.

Methodsm9d93858d85c2e49e_1528449534267_0Methods

1. Learning by observation.

2. Discussion - solving computational problems.

Forms of workm9d93858d85c2e49e_1528449514617_0Forms of work

1. Individual work.

2. Work with the whole class.

Lesson stages

Introductionm9d93858d85c2e49e_1528450127855_0Introduction

Task 1

Answer the questions:

1. Give the definition of uniform rectilinear motion.
2. Explain why velocity is a vector quantity.

Answer:

1. A uniform rectilinear motion is a motion in which the trajectory is a straight line and the speed of the moving body is constant.

2. Velocity is a vector quantity. To describe the movement of the body is not enough to give the speed with which it moves. It is important along which track and in which direction the body moves. The direction is important, i.e. the line along which the body moves in a certain direction (sense of a vector). The velocity is a vector - it has a magnitude and direction (and sense of a vector).

Procedurem9d93858d85c2e49e_1528446435040_0Procedure

[Illustration 1]

Task 2

Look at the v(t) dependence graph and answer the questions.

1. Describe how the body speed changes in this motion?

2. How would you call such a motion?

3. Is the average speed in this motion always equal to the instantaneous speed?

4. Suggest the method of determining the average speed in this motion. (Hint: think about how you calculate your average speed on the way to school).

Answer:

1. The speed in this motion increases, decreases and is constant.
2. This is a non‑uniform motion with a variable speed.
3. The instantaneous speed value is not equal to the average speed.
4. It would be necessary to calculate the area of the figure under the graph, because as we know:

v_{a v e r a g e} = \frac{s_{t o t a l}}{t_{t o t a l}}

The area of the figure below the graph is equal to the $s_{total}$ .

Definitions:

a) The motion in which the speed change occurs is called non‑uniform motionnon‑uniform motionnon‑uniform motion.
b) The motion in which the speed increases is called accelerated motionaccelerated motionaccelerated motion.
c) The motion in which the speed decreases is called decelerated motiondecelerated motiondecelerated motion.

[Slideshow]

Task 3

Questions for the slideshow:

1. What kind of motion is it?
2. What quantities describing train motion change?
3. How can we calculate accelerationaccelerationacceleration of this train?
4. In what unit we express acceleration?
5. Is acceleration a scalar or a vector quantity?

Answer:

1. It is a non‑uniform motionnon‑uniform motionnon‑uniform motion.
2. The speed changes while the time of travel increases.
3. Acceleration can be calculated by dividing the speed change by the time of this change.
4. The acceleration in the SI system has the unit $\frac{m}{s^{2}}$ .
5. Acceleration is a vector quantity, because the direction (and sense) of motion is important.

Definitional formula for acceleration:

a c c e l e r a t i o n = \frac{s p e e d c h a n g e}{t i m e o f c h a n g e}

a = \frac{Δ v}{t}

Acceleration unit in the SI system:

[\frac{\frac{m}{s}}{s}] = [\frac{m}{s} \cdot \frac{1}{s}] = [\frac{m}{s^{2}}]

Task 4

Calculate with what accelerationaccelerationacceleration the car moves, when in 5 seconds it increased its speed from 0 to 15 $\frac{m}{s}$ .

Analysis:

Given:

$v_{i n i t i a l} = 0$

$v_{f i n a l} = 15 \frac{m}{s}$

$t = 5 s$

Unknown:

$a = ?$

Solution:

$a = \frac{Δ v}{t}$

$Δ v = v_{f i n a l} - v_{i n i t i a l}$

$Δ v = 15 \frac{m}{s} - 0$

$Δ v = 15 \frac{m}{s}$

$Δ v = \frac{15 \frac{m}{s}}{5 s}$

$a = 3 \frac{m}{s^{2}}$

Answer:

The acceleration of the car is 3 $\frac{m}{s^{2}}$ .

Definition:

If the body decreases its speed, the acceleration has a negative value. We are talking about a deceleration.

Task 5

Calculate the value of the acceleration of the body, which reduces its speed from 50 $\frac{m}{s}$ to 30 $\frac{m}{s}$ in 10 seconds.

Analysis:

Given:

$v_{i n i t i a l} = 50 \frac{m}{s}$

$v_{f i n a l} = 30 \frac{m}{s}$

$t = 10 s$

Unknown:

$a = ?$

Solution:

$a = \frac{Δ v}{t}$

$Δ v = v_{f i n a l} - v_{i n i t i a l}$

$Δ v = 30 \frac{m}{s} - 50 \frac{m}{s}$

$Δ v = - 20 \frac{m}{s}$

$Δ v = \frac{- 20 \frac{m}{s}}{10 s}$

$a = - 2 \frac{m}{s^{2}}$

Answer:

Acceleration has a negative value of -2 $\frac{m}{s^{2}}$ , so we are talking about a deceleration of 2 $\frac{m}{s^{2}}$ .

Task 6

Determine which sentences are true.

Acceleration of 6 $\frac{m}{s^{2}}$ informs us that:

a) the body changed its speed by 3 $\frac{m}{s}$ during 0,5 s,
b) the body changed its speed by 6 $\frac{m}{s}$ in 1 second,
c) the body changed its speed by 1 $\frac{m}{s}$ during 6 seconds,
d) the body changed its speed by 12 $\frac{m}{s}$ in 2 seconds.

Answer:

a) True.

b) True.

c) False.

d) True.

Lesson summarym9d93858d85c2e49e_1528450119332_0Lesson summary

1. A motion in which there is a change in speed or in a speed and direction is called a non‑uniform motionnon‑uniform motionnon‑uniform motion.

2. If during the motion the speed of the body increases, we say that it is an accelerated motion, and if this speed decreases - it is a decelerated motiondecelerated motiondecelerated motion.

3. The physical quantity describing the change of the body speed in a unit of time is called acceleration. We calculate it using the formula.

a c c e l e r a t i o n = \frac{s p e e d c h a n g e}{t i m e i n t e r v a l}

a = \frac{Δ v}{Δ t}

4. AccelerationaccelerationAcceleration of one meter per second squared (1 $\frac{m}{s^{2}}$ ) informs us that the speed of the body increased by one meter per second in every second of the motion

5. Deceleration is acceleration with a negative value. Appears when the body reduces the speed of its motion (brakes).

6. In non‑uniform motion, the instantaneous velocity differs from the average velocity.

Selected words and expressions used in the lesson plan

accelerated motionaccelerated motionaccelerated motion

accelerationaccelerationacceleration

decelarationdecelarationdecelaration

decelerated motiondecelerated motiondecelerated motion

final velocityfinal velocityfinal velocity

initial velocityinitial velocityinitial velocity

non‑uniform motionnon‑uniform motionnon‑uniform motion

velocity increasevelocity increasevelocity increase

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acceleration1

non‑uniform motion1

accelerated motion1

decelerated motion1

decelaration1

final velocity1

initial velocity1

velocity increase1

Scenariusz

Topicm9d93858d85c2e49e_1528449000663_0Topic

Lesson stages

Selected words and expressions used in the lesson plan