A car in a straight line on a straight road, the first speed V1 driving two-thirds of the distance Then, one third of the distance is finished with the speed V2 = 20km / h. If the average speed of the car in the whole process is 28km / h, then V1 = () km / h

A car in a straight line on a straight road, the first speed V1 driving two-thirds of the distance Then, one third of the distance is finished with the speed V2 = 20km / h. If the average speed of the car in the whole process is 28km / h, then V1 = () km / h




Physics of grade one in Senior High School: a moves in a straight line with uniform speed V1 on a straight road
A moves in a straight line at a constant speed V1 on a straight road. At a certain time, B starts from a point O directly behind a and pursues a car at a constant speed V2 (V2 > V1). After a short period of time t, C also starts from point O and pursues a car at a constant speed v2. When B and C catch up with a, they immediately turn around and return to point O, and the speed remains the same
(1) After B overtakes a, how long does it take C to overtake a?
(2) What is the time interval between B and C returning to o?
I have solved the first question in a different way
Suppose that when B starts, the distance between a and O is x0, and the time for B to catch up with a is T1, then V2 * T1 * = V1 * T1 + x0
The time for a to catch up with B is T2, then V1 * t + x0 + V1 * T2 = V2 * T2
From the meaning of the title, △ t = T1-T2 = V1 * t / (v2-v1) (correct)
The second question is how to do it according to my method. I can't understand it


According to LZ's problem-solving method, we use distance to calculate time, so I also use distance to calculate time to do the second question. B's total journey is set as S1: S1 = 2v2 * T1, C's total journey is set as S2: S2 = 2v2 * T2, then B's total movement time is T1 = S1 / V2 = 2t1 -------- 1, C's total movement time is T2 =



A and B drive to the same target from a certain place along a straight road at the same time. A moves in a uniform straight line at the speed V1 in the first half of the time and V2 in the second half of the time (V1 ≠ V2); B moves in a uniform straight line at the speed V1 in the first half of the distance and V2 in the second half of the distance (V1 ≠ V2)
A. A car arrives first B. B car arrives first C. A and B arrive at the same time D. not sure


The distance between a and B is regarded as 1. Suppose that the time taken by a from a to B is T1, and the time taken by B from a to B is T2, then the time taken by a from a to B is T1 = 2v1 + V2, and the time taken by B from a to B is T2 = 12v1 + 12V2 = V1 + v22v1v2



The speed of a is twice that of B, and the movement time of a is three times that of B. then the distance that a passes through is ()
A. 2 times B. 3 times C. 6 times D. 5 times


∵ V A = 2V B, t a = 3T B, ∵ s a s b = v a t a v b t b = 2V B × 3T b v b t b = 6