For an object moving in a straight line, the average speed of the first half of the displacement is 3m / s, and the average speed of the second half is 2m / s. then the average speed of the whole process is? Please explain why, not just a formula, answer, I want to know how to calculate

Very simple, directly according to the definition of the expression, and then do a little deformation
If the displacement is 2S, the first half of the displacement will take T1, and the second half will take T2
S / T1 = 3, S / T2 = 2, the average speed of the whole course v = 2S / (T1 + T2);
The rest is mathematical processing: observe the known and unknown, you can find that the average velocity V can be obtained as follows
Taking the reciprocal of both sides of V = 2S / (T1 + T2) at the same time, we get 1 / v = (T1 + T2) / 2S = 0.5 [T1 / S + T2 / S] = 0.5 [1 / 3 + 1 / 2] = 0.5 [5 / 6] = 5 / 12, so we can easily get v = 12 / 5 = 2.4

RELATED INFORMATIONS

1. When an object moves in a straight line with uniform speed change, the displacement in the third second is 15 m, the displacement in the eighth second is 5 m, and the acceleration is - 2 m per square second I solved 19 meters per second, but the answer was 20 meters per second. I used the calculator to verify it countless times. I don't know what's wrong PS: the acceleration is added after I calculate it. If you have any questions, you can calculate the acceleration according to the conditions
2. In △ ABC, ∠ C = 90 ° if C = M & # 178; + n & # 178;, B = M & # 178; - N & # 178; (M > 0, n > 0), then a =?
3. In △ ABC, ∠ C = 90 ° if side length a = M & # 178; - N & # 178;, C = M & # 178; + n & # 178; (M > n > 0), then B=
4. In △ ABC, ∠ C = 90 °, if B = m, ∠ a = 30 °, then a =, C=
5. As shown in the figure, △ ABC, ab = 9, AC = 6, ad ⊥ BC at point D, M is any point on ad, then the value of mb2-mc2 is______ ．
6. In the isosceles triangle ABC, D is the midpoint of BC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively （1）△BDE≌△CDF (2) When ∠ A is equal to what degree, Quad AEDF is square? Prove your conclusion Point D is on BC, and a is the vertex angle
7. If the lengths of the three sides of △ ABC are M2-1, 2m, M2 + 1 (M > 1), then the degree of the largest angle is______ ．
8. If the three sides of triangle ABC are the square of M minus 1,2m, and the square of m plus 1, the maximum angle is
9. The two equations of X (M + 5) * 2 - (2m-5) + M-8 = 0, (1) find the value of M; (2) find the length of three sides of triangle ABC Sorry, the last question is wrong, now add. Thank you, you master! It is known that ⊿ ABC, a, B and C are the opposite sides of ⊿ A and ⊿ B ⊿ C (a > b) of triangle ABC, and a ^ 2 + B ^ 2 = C ^ 2. Sina and AINb are two of the equations (M + 5) x ^ 2 - (2m-5) x + M-8 = 0 of X respectively, (1) find the value of M; (2) find the length of three sides of ⊿ ABC
10. In △ ABC, the lengths of edges a, B and C are M2 + m + 1, 2m + 1 and M2-1 respectively. Find the maximum angle of the triangle
11. The average speed of the first step is 2m / s, and that of the second half is 3m / s. what is the average speed of the whole process
12. The initial velocity is 2 m / s and the acceleration is 1 m / S 2 What is the displacement of the particle in 5S?
13. The acceleration of an object a = 0.5m/s when it moves in a straight line with uniform acceleration at initial velocity V0 = 5m / s 1. Find the speed at the beginning of the fifth second 2. Average speed in the first three seconds
14. When a car moves at a constant acceleration of 0.5m/s2, passing through two points AB 125m apart for 10s, the speed of the car after passing through point a for 5S is 0.5m/s2
15. An object makes a uniform zero acceleration linear motion with an initial velocity of 2m / s and an acceleration of 1m / S ^ 2. Find 1. The velocity of the object at the end of 3S. 2. The object moves in the first 4S 2. The displacement of the object in the first four seconds 3. Average speed in 4S 4. The displacement of the object when the speed reaches 8m / s Ask 1.3 questions
16. An object starts from 1 m / s and rises for 3 s with uniform acceleration a, and the velocity reaches 4 m / s (1) Acceleration a of uniform acceleration rise (2) Height of rise H
17. The object moves in a straight line with a uniform velocity of 5 m / s. The displacement in the third second is 2.5 m, and the acceleration of the object is the process of explanation and calculation
18. The rated power of automobile engine is 80kW, the mass is 5T, the resistance is 0.1 times of the vehicle weight, and the acceleration at the speed of 10m / s
19. A 5T car, the rated power of the engine is 80kW, the car from static start to acceleration of 1 meter per second to do uniform acceleration linear motion, the car resistance It is 0.06 times of the vehicle weight, G is taken as 10, and the maximum time for the vehicle to make a uniform acceleration straight line? The instantaneous power at the end of 5S and 15 seconds after the vehicle starts to move?
20. The weight of a vehicle is 5T, the rated power is 80kW, the acceleration a = 1m / S2 is used for uniform acceleration linear motion, the resistance of the vehicle is 0.06 times of the vehicle weight, G is taken as 10m / S2 To find the average power of the car in the process of uniformly accelerating linear motion