Lesson plan

Topicm2fd926850dc5d4ed_1528449000663_0Topic

Study on uniform rectilinear motionuniform rectilinear motionuniform rectilinear motion

Levelm2fd926850dc5d4ed_1528449084556_0Level

Second

Core curriculumm2fd926850dc5d4ed_1528449076687_0Core curriculum

III. Planning and conducting observations or experiments and inference based on their results. The student:

II. 18) b) determines the speed based on time and distance measurement using analogue or digital instruments or software for video measurements.

Timingm2fd926850dc5d4ed_1528449068082_0Timing

45 minutes

General learning objectivesm2fd926850dc5d4ed_1528449523725_0General learning objectives

Determination of quantities describing uniform rectilinear motion.

Key competencesm2fd926850dc5d4ed_1528449552113_0Key competences

1. Recognising quantities describing the motion.

2. Experimental determination of distance and time in uniform rectilinear motion.

3. Calculation of speed based on experimental data in uniform rectilinear motion.

Operational (detailed) goalsm2fd926850dc5d4ed_1528450430307_0Operational (detailed) goals

The student:

- determines quantities such as distance and time,

- calculates speed using experimental data.

Methodsm2fd926850dc5d4ed_1528449534267_0Methods

1. Learning by experimenting.

2. Learning by solving problems.

Forms of workm2fd926850dc5d4ed_1528449514617_0Forms of work

1. Group work - experimenting work.

2. Work with the whole class - solving research problems.

Lesson stages

Introductionm2fd926850dc5d4ed_1528450127855_0Introduction

Task 1

Answer the questions:

a) What quantities are used to describe the movement of the body?

b) What do we call the motion track?

c) How do we calculate the speed of the body?

Answer:

a) The quantities used to describe the movement of the body are: the displacement of the body relative to the reference system, the time in which this displacement occurred and the path of this displacement.

b) The track of motion is the line along which the body's movement takes place.

c) We calculate the body speed from the formula:

speed = \frac{distance}{time of travel}

v = \frac{s}{t}

Procedurem2fd926850dc5d4ed_1528446435040_0Procedure

Experiment 1

Research problem:
Determining the relationship between physical quantities describing the rectilinear motion of bodies moving at a constant speed.

Hypothesis:
In uniform rectilinear motionuniform rectilinear motionuniform rectilinear motion, the distance is directly proportional to time, and the instantaneous speed is constant.

What will be need:

1) a glass tube about 1 m long and 1 cm in diameter;
2) oil, olive oil or water;
3) a ruler or paper tape (length 100 cm);
4) two plugs matched to the diameter of the glass tube;
5) stopwatch;
6) pen or pencil.

Instruction:

1) Fill the glass tube with liquid.
2) Leave a small air bubble in the tube.
3) Close both ends of the tube with plugs.
4) Attach the paper tape to the wall of the glass tube or place a ruler on it.
5) Turn the tube over and observe the movement of the air bubble.
6) Mark the position of the air bubble on a tape or ruler at regular intervals.
7) Record the results in the table below:m2fd926850dc5d4ed_1527752256679_01) Fill the glass tube with liquid.
2) Leave a small air bubble in the tube.
3) Close both ends of the tube with plugs.
4) Attach the paper tape to the wall of the glass tube or place a ruler on it.
5) Turn the tube over and observe the movement of the air bubble.
6) Mark the position of the air bubble on a tape or ruler at regular intervals.
7) Record the results in the table below:

[Table 1]

Summary:

The distances that the bubble travelled were the same at regular intervals.

Task 2

Make a distance vs. time graphgraphgraph for the results obtained as shown in illustration.

[Illustration 1]

Task 3

Make a graph of speed vs. time for this motion as shown in illustration.

[Illustration 2]

Conclusion:

In the uniform rectilinear motionuniform rectilinear motionuniform rectilinear motion, the body travels the same distances at regular time intervals.

In uniform rectilinear motion, instantaneous speed is constant. This means that the instantaneous speed in this movement is equal to the average speed. In the speed vs. time graphgraphgraph, we get a straight linestraight linestraight line parallel to the time axis.

[Interactive graphics]

Task 4

During 15 minutes, Janek travelled 1500 m and Marek at the same time covered 2 km. Calculate the speed of both boys. Which one was moving at a higher speed. Show results in $\frac{m}{s}$ and $\frac{km}{h}$ .

Solution:

Given:

t = 15 min = 15 \cdot 60 s = 900 s = \frac{1}{4} h

s_{1} = 1500 m

s_{2} = 2 k m

Unknown:

v_{1} = ?

v_{2} = ?

v_{1} = \frac{s_{1}}{t}

v_{1} = \frac{1500 m}{900 s}

v_{1} = 1, 66 (6) \frac{m}{s}

v_{1} = 1, 67 \frac{m}{s} = 1, 67 \cdot \frac{\frac{1}{1000} k m}{\frac{1}{3600} h}

v_{1} = 6, 01 \frac{k m}{h}

v_{2} = \frac{s_{2}}{t}

v_{2} = \frac{2 k m}{\frac{1}{4} h}

v_{2} = 8 \frac{k m}{h}

v_{2} = 8 \frac{k m}{h} = \frac{8 \cdot 1000 m}{3600 s}

v_{2} = 2, 22 \frac{m}{s}

Answer:

Janek moved at a speed of 1,67 $\frac{m}{s}$ (6,01 $\frac{km}{h}$ ), and Marek 2,22 $\frac{m}{s}$ (8 $\frac{km}{h}$ ). Marek moved faster.

Task 5

The car covered 180 km in 2,5 hours. What distance will the car cover when driving with the same speed for 3 hours and 20 minutes?m2fd926850dc5d4ed_1527752263647_0The car covered 180 km in 2,5 hours. What distance will the car cover when driving with the same speed for 3 hours and 20 minutes?

Solution:

Given:

t_{1} = 2, 5 h

t_{2} = 3 h 20 min = 3 \frac{1}{3} h

s_{1} = 180 k m

v_{1} = v_{2} = const

Unknown:

s_{2} = ?

v_{1} = \frac{s_{1}}{t_{1}}

v_{2} = \frac{s_{2}}{t_{2}}

\frac{s_{1}}{t_{1}} = \frac{s_{2}}{t_{2}}

s_{2} = \frac{s_{1} \cdot t_{2}}{t_{1}} [\frac{k m \cdot h}{h} = k m]

s_{2} = \frac{180 \cdot 3 \frac{1}{3}}{2, 5}

s_{2} = 240 k m

Answer:

The car will travel 240 km.

Task 6

Calculate the time it takes to travel 300 m at a speed of 20 $\frac{k m}{h}$ . Give the result in seconds, minutes and hours.

Solution:

Given:

s = 300 m = 0, 3 k m

v = 20 \frac{k m}{h}

Unknown:

t = ?

v = \frac{s}{t}

t = \frac{s}{v} [\frac{k m}{\frac{k m}{h}} = k m \cdot \frac{h}{k m} = h]

t = \frac{0, 3}{20}

t = 0, 015 h

t = 0, 015 \cdot 60 min

t = 0, 9 min

t = 0, 9 \cdot 60 s

t = 54 s

Answer:

The time required to cover the distance of 300 m is 54 s, (0,9 min, 0,015 h).

Lesson summarym2fd926850dc5d4ed_1528450119332_0Lesson summary

In order to experimentally determine the speed value, the distance and the motion time should be measured.

The speed in uniform rectilinear motionuniform rectilinear motionuniform rectilinear motion has the constant value.

The value of:

- the speed in this motion is calculated from the formula $v = \frac{s}{t}$ ,
- the distance in this motion is calculated from the formula $s = v t$ ,
- the time in this motion is calculated from the formula $t = \frac{s}{v}$ ,

where:
v - speed (e.g. $\frac{k m}{h}$ , $\frac{m}{s}$ ),
t - time (e.g. s, h),
s - distance (e.g. km, m).

The graph of the speed value vs. time is the horizontal line parallel to the time axis.

[Illutration 3]

Selected words and expressions used in the lesson plan

average velocityaverage velocityaverage velocity

graphgraphgraph

instantaneous velocityinstantaneous velocityinstantaneous velocity

straight linestraight linestraight line

uniform motionuniform motionuniform motion

uniform rectilinear motionuniform rectilinear motionuniform rectilinear motion

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uniform rectilinear motion1

graph1

straight line1

average velocity1

instantaneous velocity1

uniform motion1

Scenariusz

Topicm2fd926850dc5d4ed_1528449000663_0Topic

Lesson stages

Selected words and expressions used in the lesson plan