The speed of a train at the end of the fourth second can be calculated by making a uniformly accelerated linear motion with the initial speed of 2 meters per second and the square acceleration of 0 and 5 meters per second And the average speed of the train in the first four seconds

The speed of a train at the end of the fourth second can be calculated by making a uniformly accelerated linear motion with the initial speed of 2 meters per second and the square acceleration of 0 and 5 meters per second And the average speed of the train in the first four seconds

U (end of four seconds) = U0 + A * t = 2 + 0.5 * 4 = 4;
U (average) = s / T = (U * t + 0.5 * a * t * t) / T = 3

An object moves in a straight line with an initial velocity of 10 meters per second and an acceleration of 2 meters per second. When the velocity becomes 16 meters per second, what is the required time? What is the displacement? What is the distance of the object?

So V1 = - 16m / sv1 = V0 at = 10-2t = - 16 so t = (10 + 16) / 2 = 13s displacement x = v0t-1 / 2at ^ 2 = 10 * 13-1 / 2 * 2 * 13 ^ 2 = - 39m displacement X1 = V0 ^ 2 / (2a) = 10 ^ 2 / (2 * 2) = 25m total distance S = 2 | X1 + | x = 2 * 25 + 39

In uniform velocity linear motion, the difference of displacement in adjacent equal time is a constant
Please give an example to show that the displacement difference of adjacent equal time is a constant in the uniform linear motion

For example, free falling motion: suppose that the acceleration of gravity is 10m / S2, the first second motion distance is 5m, the second motion distance is 15m, the third motion distance is 25m, and the fourth motion distance is 35m. The nth second motion distance is "the average velocity in the nth second" (because the time is 1s), that is (the velocity in the nth-1s + the velocity in the nth second) /

There is a particle moving in a straight line with uniform velocity. The displacement of the particle passing through in two continuous equal time intervals is 24m and 64M respectively, and each time interval is 24m
4S, and acceleration

The instantaneous speed at the middle of a period of time of uniform speed change linear motion = the average speed of this period, the instantaneous speed at the midpoint of the first 4S (recorded as T1), V1 = the average speed of the first 4S = 24 / 4 = 6m / s, the instantaneous speed at the midpoint of the second 4S (recorded as T2), V2 = the average speed of the second 4S = 64 / 4 = 16m / SA = (V2 -

We define ABCD = ad BC, for example 2345 = 2 × 5-3 × 4 = 10-12 = - 2. If | x3ye | = 5, | x1y1 | = 1, then what is the value of X + y
E is 4

Classmate, I have to say that there is a mistake in your question. The first x3ye, e is 1 or some other number. I'll calculate it as 1 for the time being. If it's not for you, you can also analogize it. In fact, this question is to solve a quadratic equation of two variables. From the question, we can get: x-3y = 5x-y = 1, x = - 1, y = - 2, x + y = - 3. OK, if e

Given that a = bx2 / 3 = C + 1 / 5 = D-1 / 12, a, B, C and D are all greater than 1, can you arrange the four numbers a, B, C and D from small to large?

b＞d＞a＞c

A * 3 / 2 = b * 1 / 20 = C ﹣ 3 / 2 = D ﹣ 15 (a, B, C, D are non-zero natural numbers), and ABCD is arranged by size

A*3/2=B*1/20=C÷3/2=D÷15
Let B = 20, then = b * 1 / 20 = 1
therefore
A*3/2=B*1/20=C÷3/2=D÷15=1
have to
A=2/3 B=20 C=3/2 D=15
therefore
B>D>C>A

It is known that a = 3 / 2 = BX3 / 4 = cx11 / 12 = DX4 / 15 and ABCD is a non-zero number, please arrange ABCD from small to large
If it's urgent, I can offer a reward

Available from known sources
a=3/2
b=3/2*4/3=2
c=3/2*12/11=18/11
d=3/2*15/4=45/8
So ABCD should be C from small to large

(1) How much more is the sum of 5 out of 6 and 4 out of 7 than 5 out of 12? (2) the sum of 1 out of 3 and 2 out of 5 plus a number, the sum is 11 out of 12
Count. Help!

(5/6+4/7)-5/12
=(70/84+48/84)-35/84
=83/84
11/12-(1/3+2/5)
=55/60-(20/60+24/60)
=55/60-44/60
=15/60
=1/4

13 and 5 / 6 - (- 3 / 4) + 5 / 6 - (- 7 / 12)

13 and 5 / 6 - (- 3 / 4) + 5 / 6 - (- 7 / 12)
=13+5/6+3/4+5/6+7/12
=13+10/12+9/12+10/12+7/12
=13+36/12
=13+3
=16;
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