5. The relation between the displacement of a particle and time is: S = 4t-2t2. The units of S and T are m and s respectively,

s = 4t-2t2=4t-1/2*4*t^2
According to the formula s = v0t + 1 / 2at ^ 2:
Initial velocity V0 = 4m / S
Acceleration a = - 4m / m ^ 2 (or deceleration a = 4m / S ^ 2)

The relationship between the displacement of a particle and time is: S = 4t-2t2, the unit of S & nbsp; and t is m and s respectively, then the initial velocity and acceleration of the particle are ()
A. 4m / s and 2m / S2B. 4m / s and - 4m / S2C. 4m / s and 4m / S2D. 4m / s and 0

According to the relation between the displacement of uniform velocity motion and time x = v0t + 12at2: V0 = 4m / s, a = - 4m / S2, so B

The relationship between the displacement of a particle and time is x = 4T + 2t2. The units of X and T are m and s respectively, then the initial velocity and acceleration of the particle are respectively And ．

According to x = v0t + 12at2 = 4T + 2t2, we know the initial velocity V0 = 4m / s, acceleration a = 4m / S2. So the answer is: 4m / s, 4m / S2

Taking AB and AC of △ ABC as sides, square ABDE and ACGF are made outwards, an ⊥ BC is made at point n, and Na is extended to intersect EF at point M. the result shows that EM = MF

As shown in the figure below, ∵ EAP + ≌ ban = 90 °, ban + ≌ ABN = 90 ° and ≌ EAP = ≌ ABN. In RT △ EAP and RT △ ABN, EA = ab ≌ EAP = ≌ ABN, EPA = ≌ anb and ≌ EAP ≌ RT ≌ ANC, then FH = an = EP RT △ EMP ≌ RT △ FMH, so EM = MF

Make square ABDE and ACGF outside triangle ABC, M is the midpoint of BC, and find EF ⊥ am

Lengthen Ma, cross EF to n
ABDE and ACGF are squares
∴AB=AE,AC=AF
∠BAE=∠CAF=90°
Extend am, intercept MH = am, connect BH
∵ m is the midpoint of BC, then BM = cm,
∠BMH=∠CMA
∴△AMC≌△BMH(SAS)
∴AC=BH=AF,∠CAM=∠BHM=∠BHA
∴∠BHA+∠BAH=∠BAH+∠CAM=∠BAC
∵∠BAC+∠EAF=390°-∠BAE-∠CAF=360°-90°-90°=180°
∠ ABH + (∠ bah + ∠ BHA) = 180 ° i.e. ∠ ABH + ∠ BAC = 180 °
∴∠EAF=∠ABH
∵AB=AE,BH=AF
∴△ABH≌△EAF(SAS)
∴∠BAM=∠AEN
∵∠BAM+∠EAN=180°-∠BAE=180°-90°=90°
∴∠AEN+∠EAN=90°
Then ∠ ane = 90 °
That is am ⊥ EF

It is known that ABDE, m, N, P and Q are the middle points of EF, BC, EB and FC respectively, and mpnq is proved to be square% d% a

Connecting EC and FB, the intersection point is O, we can see that in △ FBE, MP is the median line, then there is MP / / FC and MP = FC; similarly, in △ FBC, QN is the median line, then there is QN / / FC and QN = FC. Therefore, MP and NQ are parallel and equal. Mpnq is a parallelogram, If cab is obtuse angle: ∠ Fab = 360 °- (∠ cab + ∠ CAF) = 270 °- ∠ cab, ∠ EAC = 360 °- (∠ BAC + ∠ BAE) = 270 °- ∠ cab, if cab is right angle: ∠ Fab = ∠ EAC = 180 ° in a word, ∠ Fab = ∠ EAC because AB = AE, AC = AF, it can be proved, △ Fab ≌ △ CAE, so FB = EC, so MP = MQ, It is deduced that the parallelogram mpnq is rhombic. Because of △ Fab ≌ △ CAE, so ∠ AFB = ∠ ace, there are: ∠ COF = 180 °～ FCO - ∠ CFO = 180 °～ FCA - ∠ CFA = ∠ CAF = 90 °, that is, of ⊥ OC, that is, BF ⊥ EC, so MP ⊥ MQ. It is deduced that the rhombic mpnq is square

As shown in the figure, in the regular triangular prism abc-a1b1c1, if the length of all edges is 1, then the distance from point B1 to plane ABC1 is______ ．

As shown in the figure, take AB to get the midpoint m, connect cm, C1M, pass through point C to make CD ⊥ C1M, the perpendicular foot is d ⊥ C1a = C1b, M is the midpoint of AB, ≁ C1M ⊥ ab ∩ CA = CB, M is the midpoint of AB, ∩ cm ⊥ ab ∩ c1cm = m, ∩ ab ⊥ plane c1cm ∩ ab ⊂ plane ABC1, ∩ plane ABC1 ⊥ plane ABC1 ∩ plane

As shown in the figure, the regular triangular prism abc-a1b1c1, Aa1 = AB, e is the midpoint of the side edge Aa1, (1) prove that BC1 is perpendicular to EC, (2) find the size of dihedral angle a-b-c
Can we use the liberal arts method instead of the rectangular coordinate system?

First question:
Let BC1 ∩ B1C = o
Abc-a1b1c1 is a regular triangular prism, Aa1 = AB, aa1b1b, aa1c1c and bb1c1c1c are congruent squares
The results show that Bo = C1O = b1o = Co, and BC1 ⊥ B1C ······· ①
∵ e is the midpoint of Aa1,
By Pythagorean theorem, it is easy to calculate: be = C1E, combined with the proof of Bo = C1O, we get: BC1 ⊥ EO ···································
From ①, ② and BC1 ∩ EO = O, we get: BC1 ⊥ plane eb1c, ∩ BC1 ⊥ EC
Second question: are you busy and careless? Do you want to find the dihedral angle a-be-c? If so, the method is as follows
Let the midpoint of AB be D, DF ⊥ be cross be to F, cf
∵ aa1b1b, aa1c1c, bb1c1c are congruent squares, ∵ AB = BC = AC, and D is the midpoint of ab,
∴CD⊥AB.
Obviously, there is plane ABC ⊥ plane abb1a1, and plane ABC ∩ plane abb1a1 = ab. combined with the proved CD ⊥ AB, it is obtained that:
CD ⊥ plane abb1a1, DF is the projection of CF on plane abb1a1,
By the three perpendicular theorem, there are: CF ⊥ be
From DF ⊥ be and CF ⊥ be, it is concluded that ∠ CFD is the plane angle of dihedral angle a-be-c
∵ RT △ Abe and RT △ FBD have a common acute angle, that is ∠ Abe, ∵ RT △ Abe ∵ RT △ FBD,
∴DF/AE＝BD/BE,∴DF＝BD×AE/BE.
∴tan∠CFD＝CD/DF＝CD×BE/（BD×AE）
＝（√3/2）AB√（AB^2＋AE^2）/［（1/2）AB×（1/2）AA1］
＝2√3×√［AB^2＋（AA1/2）^2］/AB
＝2√3×√［AB^2＋（AB/2）^2］/AB
＝2√3×√（1＋1/4）
＝√15.
The dihedral angle a-be-c is arctan √ 15
Note: if the second question is not what I guessed, please add

The mathematics questions take the sides AB and AC of the triangle ABC as the sides to make the square ABDE and the square acfg outward respectively, connect the eg, and try to judge the triangle ABC and the triangle AE
Take sides AB and AC of triangle ABC as sides, make square ABDE and square acfg outward respectively, connect eg, try to judge the relationship between triangle ABC and triangle AEG area, and explain the reason
No sine function

Equal
Area of △ ABC = 1 / 2 * AB * ac * sin ∠ bac
Area of △ AEG = 1 / 2 * AE * Ag * sin ∠ EAG
AB=AE AC=AG ∠BAC=180º-∠EAG
So the area of △ ABC = △ AEG

In the triangle ABC, take AB and AC as sides, make square ABDE and square acfg outward respectively, connect eg, M is the midpoint of BC, and find eg = 2am

Extend am to P so that am = MP
The triangle AMB and PMC are congruent, so AB is parallel to PC
Angle EAG + BAC = 180 'angle BAC + ACP = 180', so angle EAG equals angle ACP, AE equals CP, Ag equals AC, triangle EGA is congruent PAC, so am equals half of eg

5. The relation between the displacement of a particle and time is: S = 4t-2t2. The units of S and T are m and s respectively,