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A 200m long train passed the bridge. It was measured that it took 1 minute for the train to get off the bridge. The duration of the whole train on the bridge was 40s. Calculate the length of the bridge and the speed of the train
If the bridge length is x and the train speed is y, then (1min is converted to 60s) X-200 = 40y (the whole train is completely on the bridge for 40s, which refers to the period from the beginning of the bridge at the rear of the train to the end of the bridge at the front of the train. When the bridge is on the rear of the train, the front of the train has been on the bridge for 200m, so the distance traveled by the train is the bridge length - vehicle length.) x + 200 = 60y (the train starts from
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 = 12a1t2 ① Displacement in deceleration stage: x2 = v0t − 12a2t2 ② Where: V0 = a1t x1=-x...
Part of a section of the road is uniformly accelerated, and the acceleration is A1, and then uniformly decelerated, and the acceleration is A2. The total time of this section is s
Let the time of accelerating motion be T1, the time of decelerating motion be T2, total time t = T1 + T2, initial speed of accelerating motion = final speed of decelerating motion = 0, A1 * T1 = A2 * t2t1 / T2 = A2 / A1 (T1 + T2) / T2 = (a1 + A2) / a1t / T2 = (a1 + A2) / a1t = [(a1 + A2) / A1] T2. Average speed v '= A2 * T2 / 2 = s / TT2 = 2S / (t * A2)
An object makes a uniform acceleration with the acceleration of A1, and then makes a uniform deceleration with the acceleration of A2. T time stops and asks the displacement
Suppose the maximum speed of the object is V, the time for uniform acceleration is T1, the time for uniform deceleration is T2, and the total displacement is s
S=1/2*a1*t1*t1 + 1/2*a2*t2*t2
Note that A1 * T1 = A2 * T2 = V, T1 + T2 = t, so:
S=1/2*V（t1+t2)=1/2*VT
T is known and V is required
According to:
t1+t2=T
a1*t1-a2*t2=0
Among them, A1, A2 and T are known, T1 and T2 are unknown. This is a binary quadratic equation, which can be obtained:
t1=a2*T/(a1+a2),t2=a1*T/(a1+a2)
Then: v = A1 * T1 = A1 * A2 * t / (a1 + A2)
Finally:
S=a1*a2*T^2/2(a1+a2)
2. If the initial velocity V0 = 2m / s and the acceleration a = 1m / S2 at the beginning of the uniformly accelerated linear motion of an object, what is the velocity M / s at the end of the third second If the acceleration is a = - 1m / / s, what is the speed after 2 seconds
(1) V (end of 3S) = V0 + at = 2 + 1 × 3=5m/s
(2) V (end of 2S) = V0 + at = 2 + (- 1) × 2=0m/s
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 the particle just returns to the original starting point after time t, A1 ∶ A2 should be
S1=0.5*a1*t^2
v1=a1*t
t2=v1/a2
S2=0.5*a2*t2^2
S3=0.5*a2*(t-t2)^2
With the formula S1 + S2 = S3
a1：a2=1：3
RELATED INFORMATIONS
- 1. An object starts to move in a straight line with uniform acceleration from rest. The acceleration is A1. The movement time is T1. Then it moves in a uniform deceleration with acceleration A2. After T2, the speed is exactly zero. Then the average speed of the object in the whole process can be expressed as? Why are 0.5a1t1 and 0.5a2t2 also right
- 2. 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
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