The relationship between the displacement of a particle moving in a straight line and time is x = 4t-2x ^ 2. The units of X and T are m and s respectively

The relationship between the displacement of a particle moving in a straight line and time is x = 4t-2x ^ 2. The units of X and T are m and s respectively

Compared with the displacement formula, s = v0t + 1 / 2at ^ 2, it can be concluded that
The coefficient of T is V0, and the coefficient of T ^ 2 is 1 / 2A,
It's impossible to use derivation in the first year of high school. Should x on the right side of the title expression be t~

What is the displacement of the object in the sixth second when it starts to move at a constant acceleration of 2 m / s Square from rest?

Calculation process: Formula: S = v.t + 1 / 2at & sup2;
Because: v. = 0
So: S = 1 / 2 * 2 * 6 & sup2; = 36m

If an object moves with constant acceleration on a horizontal plane, the relation between its displacement and time is x = 24t-6t2, then the time when its velocity is zero is ()
A. 16S & nbsp; end B. 2S end C. 6S end D. 24s end

According to x = v0t + 12at2 = 24t − 6t2, initial velocity V0 = 24m / s, acceleration a = - 12m / S2, according to the velocity time formula, v = V0 + at solution t = 0 − v0a = 0 − 24 − 12 = 2S. So B is correct, a, C and D are wrong

The acceleration is a = 0.1M per square second. The initial velocity is v 0 = 2m per second

S=V0t+1/2at^2=2*4+1/2*0.1*4^2=8.8m
V_ =S/t=2.2m/s

As shown in the figure, take the sides AB and AC of △ ABC as the sides, make square ABDE and acfg outward respectively, connect eg, try to judge the relationship between △ ABC and △ AEG area, and explain the reason

If the area of △ ABC and △ AEG is equal, passing point C is cm ⊥ AB in M, passing point G is GN ⊥ EA, intersecting EA extension line in N, then ⊥ AMC = ⊥ ang = 90 °, ⊥ Quad ABDE and Quad acfg are square, so ⊥ BAE = ⊥ CAG = 90 °, AC = AG, ⊥ EAB ⊥ GAC = 180 °⊥ BAC ⊥ EAG = 180 °⊥ EAG ⊥ Gan

Taking the sides AB and AC of △ ABC as the sides, square ABDE and square acfg are made towards the outside of the triangle, and point m is the midpoint of BC

Prove that: extend am so that Mn = am, connect CN, extend Ma and eg to intersect at point H
Because m is the midpoint of BC
So BM = MC
Because angle AMB = angle CMN
So triangle AMB and triangle NMC congruence (SAS)
So AB = CN
Angle ABM = angle NCM
So AB is parallel to CN
So angle BAC + angle ACN = 180 degrees
Because the quadrilateral ABDE is a square
So AB = AE
Angle BAE = 90 degrees
Because the quadrilateral acfg is a square
So AC = AG
Angle CAG = 90 degrees
So CN = AE
Because angle BAE + angle BAC + angle CAG + angle EAG = 360 degrees
So angle BAC + angle EAG = 180 degrees
So angle ACN = angle EAG
So triangle ACN and triangle gae congruence (SAS)
So angle age = angle can
Because angle can + angle CAG + angle GAH = 180 degrees
So angle GAH + angle age = 90 degrees
Because angle GAH + angle age + angle AHG = 180 degrees
So the angle AHG = 90 degrees
So am vertical eg

Taking AB and AC of triangle ABC as sides, making square ABDE and acfg outward, connecting BG, CE and eg, the area of triangle ABC and triangle AEG is equal

The quadrilateral ABDE and acfg are square
∴AE=AB AC=AG
And ∵ ∠ EAG = ∠ BAC (equal to vertex angle)
∴△EAG≌△ABC
∴S△EAD=S△ABC

The edge of △ ABC AB.AC The relationship between the area of △ ABC and △ AEG can be judged by making square ABCD and square acfg for the edge outward respectively and connecting eg
explain

I've done this question before, and you can search it on the Internet. You'll be surprised. It's so similar,
Someone copies the answer and then answers you. Now what do you choose?
The area of triangle ABC is equal to that of triangle AEG. Make cm vertical AB to m, make GN vertical EA, intersect EA extension line to N, angle AMC = angle ang = 90 °,
Because both ABDE and acfg are square, the angle BAE = angle CAG = 90 °, AC = AG, angle EAB + angle GAC = 180 °
So the angle BAC + angle EAG = 180 degrees
Because angle EAG + angle Gan = 180 degrees,
So angle BAC = angle Gan,
So triangle ACM congruent triangle AGN
So cm = GN
Because AE = AB, s triangle ABC = 1 / 2 * AB * cm, s triangle AEG = 1 / 2ae * CN
So s triangle ABC = s triangle AEG
My possibility is hard to understand, but I seriously answer. Unlike some people

As shown in the figure, in the four prism abcd-a1b1c1d1, the bottom surface ABCD is a square, the side edge A1A ⊥ the bottom surface ABCD, and E is the midpoint of A1A

Prove: connect AC, let AC ∩ BD = f, connect EF, because the bottom ABCD is a square, so f is the midpoint of AC, and E is the midpoint of A1A, so EF is the median line of △ a1ac, so EF ∥ A1C. Because EF ⊂ plane EBD, A1C ⊄ plane EBD, so A1C ∥ plane EBD

In the square prism abcd-a1b2c3d4, Aa1 = 2Ab = 4, point E is on CC1 and C1E = 3ec (1) prove the A1C ⊥ plane bed (2) find the size of dihedral angle a1-de-b

It is proved that the intersection of AC and BD at point m and me is the intersection of plane acc1a1 and plane BDE. If A1C passes through plane DEB and intersects at point F, then f must be on the intersection me
(plane ACD and straight line BD)
⊙ AC ⊥ BD, Aa1 ⊥ planar ABCD = > Aa1 ⊥ BD
AA1∩AC=A
⊥ BD ⊥ plane aa1c ⊥ BD ⊥ A1C
(in plane acc1a1, RT △ MCE and RT △ aa1c)
CE：AC=1：2√2=CM：AA1=√2:4
So △ MCE ∽ aa1c = > CME = ∠ aa1c ∠ MEC = ∠ CME
So ∠ CFM = ∠ caa1 = 90 ° that is A1C ⊥ me
BD ∩ me = m so A1C ⊥ plane BDE
（2）
A1c ⊥ de from A1C ⊥ plane BDE
Make vertical intersection of de through F, then FP ⊥ de a1f ⊥ de a1f ∩ FP = f to get vertical plane a1fp of de
So de ⊥ a1p
Then, fpa1 is the dihedral angle
(find a1f length)
In plane acc1a1
△ CFE∽△CC1A1 A1C=√（（2√2）²+4²）=2√6
Let CF be X
Then x: 4 = 1:2 √ 6, the solution is x = √ 6 / 3
So a1f = a1c-cf = 2 √ 6 - √ 6 / 3 = 5 √ 6 / 3
(find pf length)
Delta BDE is an isosceles triangle, M is the midpoint, so ∠ EMD = ∠ FPE = 90 °
Therefore, △ PEF ∽ DME
PF：DM=FE：DE DE=√（1+2² ）=√5
The length of Fe in △ CEF is √ (1 & sup2; - (√ 6 / 3) & sup2;) = √ 3 / 3
DM = √ 2, PF = √ 30 / 15
(find the length of a1p)
A1p is the high value of △ a1de
DE=√5 DA1=√（2²+4²）=2√5 A1E=√（C1E²+A1C²）=√（（2√2）²+3²）=√17
From cosine theorem
Cos∠EDA1=（（√5 ）²+（2√5）²- （√17）²）/（2×√5×2√5）=2/5
So sin ∠ eda1 = √ (1 - (2 / 5) & sup2;) = √ 21 / 5
S△A1DE=1/2×A1D×DE×sin∠EDA1=1/2×DE×A1P
The solution is a1p = 2 √ 105 / 5
(angle fpa1)
In △ a1pf, cos ∠ fpa1 = ((√ 30 / 15) & sup2; + (2 √ 105 / 5) & sup2; - (5 √ 6 / 3) & sup2;) / (2 × √ 30 / 15 × 2 √ 105 / 5) = √ 14 / 42
So the dihedral angle is arccos √ 14 / 42