How to calculate the linear motion and force with uniform speed
The motion along a straight line with constant acceleration is called uniform linear motion. Although the velocity of the object changes with time, the magnitude and direction of the acceleration of the object do not change with time. An object moving in a straight line at a constant speed keeps the state of constant speed. Perhaps "uniform speed change" can also be called a state of motion. Why does an object moving in a straight line with constant speed change keep constant speed change? It's because of the force. It can be said that the force keeps an object moving in a straight line at a constant speed. More succinctly, the force keeps the object constant. The so-called constant velocity means that the acceleration is constant. Maybe we can say that the force has the property of keeping the acceleration constant
Uniform linear motion is the simplest mechanical motion, which refers to the motion along a straight line with constant speed. Today, Xiao Bian sorted out the knowledge points related to uniform linear motion, hoping to help you learn
Concept: the displacement of objects is equal in any equal time. It is a linear motion with constant velocity and zero acceleration Formula: v = s / T deformation formula: S = VT, t = s / V2. Image: S-T image of uniform linear motion is a straight line: the slope of the line is numerically equal to the velocity of the object
In the following statements, the correct one is ()A. when an object moves in a certain direction along a straight line, the distance it passes is displacement
B. when an object moves in a certain direction along a straight line, the distance it passes is equal to the displacement
C. when an object moves in a certain direction, the distance it passes is displacement, The displacement may be zero
D. the distance that the object passes through may be different, but the displacement may be the same
analysis: the displacement is a vector, and the distance is a scalar, so it can not be said that the scalar is the vector, so a is wrong, B is correct. The distance is the actual length of the object's motion trajectory, and the displacement is a directed line segment from the starting position to the ending position of the object's motion, If an object moves in a one-way straight line, the distance is equal to the displacement. If an object moves along different tracks between two positions, their displacement is the same, and the distance may be different. If an object moves from a certain position and returns to the starting position after a period of time, the displacement is zero, but the distance is not zero
answer: CD
example 2: about the relationship between speed and acceleration, the correct one in the following statement is ()
a. the greater the change of speed, the greater the acceleration
B. The faster the change of speed, the greater the acceleration
C. the size of acceleration remains unchanged, and the direction of speed also remains unchanged
C. the size of acceleration keeps decreasing, and the size of speed also keeps decreasing
analysis: according to, Δ The greater V is, the greater the acceleration is not necessarily. The faster the velocity changes, the greater the acceleration is. Therefore, the greater the acceleration is. B is correct. There is no direct connection between the acceleration and the direction of velocity. If the acceleration does not change, the direction of velocity may not change or change. If the acceleration decreases, the velocity may increase continuously.
answer: B
example 3: in a uniform electric field parallel to the x-axis, the acceleration is not directly related to the direction of velocity, The field strength is e = 1.0 × 106v / m, one charge q = 1.0 × 10-8c, mass m = 2.5 × The relation between displacement and time is x = 5-2t, where x is in M and t is in S. The distance from the beginning to the end of 5S is_________ m. The displacement is__________ m
It should be noted that the first question is the distance; The second question is displacement Compared with x = 5-2t, the initial position of the object is x0 = 5m, the initial velocity is V0 = m / s, and the movement direction is opposite to the positive direction of displacement, that is, along the negative direction of X axisanswer: therefore, the distance from the beginning of motion to the end of 5S is 10m, and the displacement is m
Case 4: a yacht glides in a straight line at a constant speed and sails against the river. It loses a life buoy somewhere. It takes T seconds to find out after the loss, so the yacht immediately goes back to catch up with it. As a result, it catches up with the yacht at s meters downstream of the loss point and calculates the water speed. (the water speed is constant, and the yacht's speed of rowing back and forth is constant)analysis: with water as the reference (or lifebuoy as the reference), the displacement of the yacht relative to the lifebuoy is the same, and the speed of the yacht relative to the lifebuoy is also the same, so the time to return to catch up with the lifebuoy is also T seconds, and the time from loss to catch up is 2T seconds, The lifebuoy moves s meters with the water.
answer: water speed v = 2S / T.
thinking: if the people on the yacht find that they are lost, the lifebuoy is s meters away from the yacht. At this time, they immediately return to catch up. It takes T seconds to catch up and find the speed
for more information, we can pay attention to the uniform linear motion formula of scientific high score network
as for the second and the end of the second, you can draw a time axis marked with 123456... Then note that the second is the corresponding number (that is, the time), and then the end of the second, such as the end of the third second, That is, infinitely close to 4 seconds.
then VT diagram, if it is a straight line parallel to the T axis, it is a uniform linear motion, because the speed does not change; If it is an oblique line, it is a straight line with uniform speed change.
if it is an oblique upward line, it is a straight line with uniform acceleration; if it is an oblique downward line, it is a straight line with uniform deceleration
1) uniformly variable speed linear motion
1. Average velocity V flat = s / T (definition)
2. Useful inference vt2-vo2 = 2As
3. Velocity V T / 2 = V flat = (VT + VO) / 2
4. Final velocity V T = VO + at
5. Velocity V S / 22 = (VO 2 + vt2) / 2
6. Displacement S = V flat t = VO T + at 2 / 2 = V T / 2
7. Acceleration a = (VT VO) / T {with VO as the positive direction, a and VO in the same direction (acceleration) a & gt; 0 In reverse, a & lt; 0}
8 Δ s＝aT2 ｛ Δ Main physical quantity and unit: initial velocity (VO): M / S; Acceleration (a): M / S2; Final velocity (VT): M / S; Time (T) seconds (s); Displacement (s): m; Distance: meters; Speed unit conversion: 1m / S = 3.6km/h
note:
(1) average speed is vector
(2) the acceleration is not necessarily large when the velocity is large
(3) a = (VT VO) / T is only a measure, not a determinant
(4) other related contents: particle, displacement and distance, reference frame, time and time [see Volume I P19] / S-T diagram, V-T diagram / velocity and velocity, instantaneous velocity [see Volume I p24]
2) free falling body motion
1. Initial velocity VO = 0
2. Final velocity VT = GT
3. Falling height h = GT2 / 2 (calculated from VO position downward)
4. Dection vt2 = 2GH
note:
(1) free falling body motion is a uniform acceleration linear motion with zero initial velocity and follows the law of uniform velocity variation linear motion
(2) a = g = 9.8 m / S2 ≈ 10 m / S2 (the acceleration of gravity is smaller near the equator, and it is smaller in the high mountains than in the plain, and the direction is vertical and downward)<
(3) vertical throwing motion
1. Displacement S = vot-gt2 / 2
2. Final velocity VT = VO GT (g = 9.8m / S2 ≈ 10m / S2)
3. Useful inference vt2-vo2 = - 2GS
4. Maximum rising height HM = VO2 / 2g (from throwing point)
5. Round trip time t = 2VO / g (time from throwing to original position)
note:
(1) whole process processing: uniform deceleration The linear motion is positive in upward direction, and the acceleration is negative
(2) segmented processing: upward is uniform deceleration linear motion, downward is free falling motion, with symmetry
(3) the rising and falling processes are symmetrical, such as the velocity equivalent at the same point and the opposite direction.
1. Average velocity VT = s / T (definition)
2. Useful inference vt2-vo2 = 2As
3. Middle time velocity VT / 2 = VT + VO) / 2
4. Final velocity VT = VO + at
5. Middle position velocity vs / 22 = (VO2 + vt2) / 2
6. Displacement S = VT, t = VT + at2 / 2 = VT / 2 T
7. Acceleration a = (VT VO) / T {with VO as positive value} Direction, a and VO are in the same direction (acceleration) a & gt; 0 In reverse, a & lt; 0}
8 Δ s＝aT2 ｛ Δ Main physical quantity and unit: initial velocity (VO): M / S; Acceleration (a): M / S2; Final velocity (VT): M / S; Time (T) seconds (s); Displacement (s): m; Distance: meters; Speed unit conversion: 1m / S = 3.6km/h
note:
(1) the average speed is a vector
(2) the acceleration is not necessarily large when the velocity is large
(3) a = (VT VO) / T is only a measure, not a determinant
(4) other related contents: particle, displacement and distance, reference frame, time and time [see Volume I P19] / S-T diagram, V-T diagram / velocity and velocity, instantaneous velocity [see Volume I p24]
2) free falling body motion
1. Initial velocity VO = 0
2. Final velocity VT = GT
3. Falling height h = GT2 / 2 (calculated from VO position downward)
4. Dection vt2 = 2GH
note:
(1) free falling body motion is a uniformly accelerating linear motion with zero initial velocity and follows the law of uniformly variable speed linear motion
(2) a = g = 9.8 m / S2 ≈ 10 m / S2 (the acceleration of gravity is smaller near the equator, and it is smaller in the high mountains than in the plain, and the direction is vertical and downward)
3) vertical throwing motion
1. Displacement S = vot-gt2 / 2
2. Final velocity VT = VO GT (g = 9.8m / S2 ≈ 10m / S2)
3. Useful inference vt2-vo2 = - 2GS
4. Maximum rising height HM = VO2 / 2g (from throwing point)
5. Round trip time t = 2VO / g (time from throwing to original position)
note: (1) whole process processing: it is uniform deceleration linear motion, Taking upward as positive direction, the acceleration takes negative value
(2) segmented processing: upward is uniform deceleration linear motion, downward is free falling motion, with symmetry
(3) the rising and falling processes are symmetrical, such as the velocity equivalent at the same point and the opposite direction.
The first one is the first one; V = V0 + at is best understood from the definition of a - the change of velocity per unit time
the original speed of the object is: V0, the acceleration of the uniform variable speed motion is a, that is, the change of the speed per second, then the change of the speed after T seconds is: at. The original speed plus the changed speed is the later speed, so: v = V0 + at
the second formula: x = v0t + (1 / 2) at ^ 2 the textbook is derived by the graphic method (speed - area in the time image)! I won't repeat it. This paper analyzes theoretically:
displacement = average velocity * time
when the initial velocity is V0
T seconds, the velocity v = V0 + at
so the average velocity V & # 39= V0 + V) / 2
so the displacement = average velocity * time
x = (V0 + V) / 2 * t = (V0 + V0 + at) / 2 * t
x = v0t + (1 / 2) at ^ 2
is a uniform linear motion, and the falling stage is a free falling body motion
If an object moves from the initial position to the end position in a certain period of time, the directed line segment from the initial position to the end position is called displacement. Its size is the linear distance from the initial position to the final position of the moving object; The direction is from the beginning to the end
The displacement is only related to the beginning and end position of the object, but not to the trajectory of the object. If the particle returns to its original position after a period of time, the distance is not zero and the displacement is zeroΔ X = x2-x1 (final position minus initial position) it should be noted that the displacement is a linear distance, not a distance
in the international system of units (SI), the main unit of displacement is meter. In addition: centimeter, kilometer and so on. The displacement formula of uniform variable velocity motion: x = v0t + 1 / 2 · at ^ 2
dection of velocity and displacement of uniform variable velocity motion: x = VOT + & # 189; at²
Note: V0 is the initial velocity and VT is the final velocity2, S1 + S2 = V0 * (2 * t) + 1 / 2 * a * (2 * t) ^ 2: 32 = 4v0 + 8A< The results show that: a = - 1, V0 = 8
from vt = V0 + A * t... 0 = 8 + (- 1) * t: T = 8 (seconds)
so the acceleration is - 1, and the particle stops after 8 seconds
question 2: V A * t = v b 0 * t + 1 / 2 * a * T ^ 2, fill in the number: 100 = v b 0 * 10 + 50A
300 = v b 0 * 20 + 200A
get: a = 1, v b 0 = 5 (M / s)
the speed of B catching up with a is: v = V0 + at = 5 + 1 * 10 = 15 (M / s)
the acceleration of B is: a = 1
question 3:
suppose the initial velocity is v0, The intermediate velocity is V, and the final velocity is vt; The first half time was T1, the second half
was T2, and the position of the front and back half was s
s = 3 * T1 = 6 * T2 leads to T1 = 2 * T2
3 = (V0 + V in) / 2 V0 = V in - A * T1, we get - A * T1 = 2V in - 2A * T2 = 6
6 = (VT + V middle) / 2 VT = V middle + A * T2, which leads to 2V middle + A * T2 = 12
it is concluded that v = 5 (M / s)
the velocity through the middle position is 5m / s