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A particle starts to move in a straight line with uniform acceleration from rest, and the acceleration is A1. After time t, it moves in a straight line with uniform deceleration, and the acceleration is A2. If it can just return to the starting point after time t, A1: A2 should be () A. 1：1 B. 1：2 C. 1：3 D. 1：4
In the acceleration phase: X1 = 1
2a1t2 ①
Displacement in deceleration stage: x2 = v0t − 1
2a2t2 ②
Where: V0 = a1t x1=-x2 ③
Simultaneous ① ② ③ solution: A1: A2 = 1:3, so abd is wrong and C is correct
Therefore, C
The car starts to move in a straight line with uniform acceleration at the acceleration of A1 from standstill. After a period of time, it moves in a uniform and reduced motion at the acceleration of A2. The whole process time is t. please write down the detailed problem-solving process. It is also useful. There is no need to discuss it by classification. Is A2 ten size rather than direction?
No discussion
When the total time is determined, it's OK,
The whole process is from the beginning to the end
The time to change the acceleration can be calculated. At that point, the speed is the same. Let it be time T1
a1*t1=a2*(t-t1)
t1=a2*t/(a1+a2)
Total displacement S = 0.5a1t1 ^ 2 + 0.5a2 (t-t1) ^ 2
The car starts to move in a straight line with uniform acceleration at the acceleration A1 from standstill, and after a period of time, it moves at the acceleration of A2 The car starts to move in a straight line with uniform acceleration at the acceleration of A1 from standstill. After a period of time, it moves in a straight line with uniform deceleration at the acceleration of A2. It stops after a total distance of L. calculate the total time of car movement There must be a process. Thank you
Let the time of the first two sections be T1 and T2 respectively, and the distance traveled is S1 and S2. Let the maximum speed reached be VT, so the initial speed in T1 is 0 and the final speed is VT, so A1 times T1 = vt.1
In T2 time, the initial speed is VT and the final speed is 0, so
Vt + A2 times T2 = 0 (Note: A2 is negative at this time, due to deceleration). 2
And S1 = [A1 times (T1) square] / 2.3
S2 = VT times T2 + [A2 times (T2) square] / 2.4
S1+S2=L.5
Connect 1 2 3 4 5 equation
It is known that the subway train runs every 10 minutes and stops at the station for 1 minute. The probability of passengers arriving at the platform to get on the train immediately is calculated The answer is one tenth or one eleventh. Should the 10 minutes include one minute to reach the destination
The answer is one tenth, because the subway train has been set to run every 10 minutes, that is to say, when the first train arrives at the station, the next train will arrive at the station exactly 10 minutes later, that is to say, in these 10 minutes, there is a train stopping at the station for 1 minute, and it can be calculated that in any 10 minutes
If the maximum acceleration that passengers can bear in the subway train is 1.4 and the distance between two adjacent stations is 560, the subway train runs between these two stations for at least 30 minutes
To minimize the time, accelerate and decelerate directly, and the acceleration is 1.4m/s2
The equation time is t, and the acceleration and deceleration time are the same, both of which are t / 2
So 1 / 2 * 1.4 * (T / 2) ^ 2 * 2 = 560 (equivalent to finding the area of a triangle)
So t = 20
The maximum acceleration that passengers can endure in the subway train is 1.4m/s ^ 2. The distance between the two necessary stops is 2240m. If the maximum running speed of the train is 28m / s, How long does the train last at these two stations? Assuming that there is no maximum speed limit, what is the shortest time for the train to travel at these two stations?
(1) When the train accelerates to the maximum running speed of 28m / s, the service time shall be at least 20s. During this period, the travel distance: S1 = 1 / 2 * 20s * 28m / S = 140m. When the train decelerates from the maximum running speed of 28m / s to stop, the service time shall be at least 20s. Similarly, during this period, the travel distance: S2 = 1 / 2 * 20s * 28m / S = 140m, the rest in the middle
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