Equitable movement

Содержание и основные понятия

The acceleration vector is constant in the same speed, the velocity changes linearly over time, and the coordinates change according to square law.
GEF · Grade 9
PIF · Baseline
Interactive model
acceleration
instantaneous speed
Time schedule
stopping distance
01
The physical picture
What happens and what boundaries the model has.
Main message
The acceleration vector is constant in the same speed, the velocity changes linearly over time, and the coordinates change according to square law.
The acceleration sign itself does not show acceleration or braking: it is compared to the speed mark.
The area under the v(t) schedule is equal to displacement.
The slope of the graph v(t) is numerically equal to acceleration.
How to reason
First, name the system and the interaction, then select the values, and only then write down the law. The formula works within the model's assumptions.
02
Laws and values
Symbols should be read as a relationship of values, not as a recipe for substitution.
v
=
v
0
+
a
t
v=v_0+at
v
=
v
0
​
+
a
t
speed
x
=
x
0
+
v
0
t
+
a
t
2
2
x=x_0+v_0t+\frac{at^2}{2}
x
=
x
0
​
+
v
0
​
t
+
2
a
t
2
​
coordinate
Review of the situation
The vehicle with v0 = 10 m/s and a = 2 m/sec2 after 5 s has a speed of 20 m/second and passes through 75 m.
04
Links and application
In the real world
The vehicle with v0 = 10 m/s and a = 2 m/sec2 after 5 s has a speed of 20 m/second and passes through 75 m.
Model verification
Compare the prediction with observation, evaluate the units, the order of magnitude and the influence of the factors not taken into account. Disparity does not always mean an error of law: often idealization is no longer sufficient.
C
What needs to be understood
Outcome 1
To explain the physical meaning of the theme "Equating motion" through the model, not just through the formula.
Outcome 2
Clear and correctly use terms: acceleration, instantaneous speed, time schedule.
Outcome 3
Indicate the conditions for the applicability of the law, verify units and limit cases.
T
Terms of the topic
The definition should indicate the physical meaning, not a substitute for a domestic synonym.
acceleration
Vector value of speed change per unit time.
instantaneous speed
Average speed limit when the time interval is zeroed; directed at the tangent to the trajectory.
Time schedule
Graphical representation of the relationship between coordinates, path or speed from time to time.
stopping distance
The distance travelled by the body from the beginning of braking to the stop.
F
Origin and conclusion of the formulas
The fundamental law cannot be "proven" by algebras: for each record, it is stated whether it is a definition, an experimental law, or a model effect.
v
=
v
0
+
a
t
v=v_0+at
v
=
v
0
​
+
a
t
model effect
Speed.
The following is an indication of the definition, experience or law from which this entry arises and where it is applicable.
a
=
v
−
v
0
t
a=\frac{v-v_0}{t}
a
=
t
v
−
v
0
​
​
At a constant acceleration, its projection is equal to the change in the projection of speed over time t.
a
t
=
v
−
v
0
at=v-v_0
a
t
=
v
−
v
0
​
The equation is multiplied over time; the value at is the accumulated change in speed.
v
=
v
0
+
a
t
v=v_0+at
v
=
v
0
​
+
a
t
Transfer the initial speed to the right; the signs v_0 and a are determined by the chosen direction of the axle.
Conditions:
Direct motion with constant projection of acceleration; t is calculated from the time when the speed is v_0.
Monitoring:
The dimensions of both parts shall coincide; the result shall be checked for sign, order and limit case.
x
=
x
0
+
v
0
t
+
a
t
2
2
x=x_0+v_0t+\frac{at^2}{2}
x
=
x
0
​
+
v
0
​
t
+
2
a
t
2
​
model effect
Coordinate.
The following is an indication of the definition, experience or law from which this entry arises and where it is applicable.
v
(
t
)
=
v
0
+
a
t
v(t)=v_0+at
v
(
t
)
=
v
0
​
+
a
t
With constant acceleration, the speed schedule is straight from time to time.
Δ
x
=
v
0
t
+
(
v
−
v
0
)
t
2
=
v
0
t
+
a
t
2
2
\Delta x=v_0t+\frac{(v-v_0)t}{2}=v_0t+\frac{at^2}{2}
Δ
x
=
v
0
​
t
+
2
(
v
−
v
0
​
)
t
​
=
v
0
​
t
+
2
a
t
2
​
The displacement is equal to the area below the speed schedule: the area of the rectangle v_0t and the triangle with height at.
x
=
x
0
+
Δ
x
=
x
0
+
v
0
t
+
a
t
2
2
x=x_0+\Delta x=x_0+v_0t+\frac{at^2}{2}
x
=
x
0
​
+
Δ
x
=
x
0
​
+
v
0
​
t
+
2
a
t
2
​
The initial coordinates are added to the displacement projection found.
Conditions:
The movement along the axle with constant acceleration a; x_0 and v_0 refer to t=0, resistance and other causes of change a are not taken into account.
Monitoring:
The dimensions of both parts shall coincide; the result shall be checked for sign, order and limit case.
!
Model boundaries and typical errors
Error: Forget model condition
The key condition of this topic is that the acceleration sign does not in itself show acceleration or braking: it is compared to the speed mark.
How to Check the Mind
The area under the v(t) schedule is equal to displacement. Then check the units, vector direction, and result behaviour at zero or very large parameter.
Σ
Conclusion on the topic
Equitable movement
The acceleration vector is constant in the same speed, the velocity changes linearly over time, and the coordinates change according to square law.
The acceleration sign itself does not show acceleration or braking: it is compared to the speed mark.
The area under the v(t) schedule is equal to displacement.
Any numerical result shall be accompanied by a unit, a sign and an explanation of its physical meaning.
The content is consistent with the FRP on Physics of the OS/OSS 2025 and the subject matter of the current lines of the PNP: I. M. Peryškin and co-authors (7–9), G. J. Mäkishev and coauthors (10–11), and with the lines "Engineers of the Future" and V. A. Kasyanov. The original texts are not reproduced, the text of the textbooks is not reproduced.