Some students took the power train to travel. When the train accelerated evenly on a straight track, one student proposed, "can we use the equipment around us to measure the acceleration of the train?" many students participated in the measurement. The measurement process was as follows: they looked out of the window every 100m   While recording the time with a watch, they observed that the time interval from the first landmark to the second landmark is 5S, and the time interval from the first landmark to the third landmark is 9s. Please find out according to their measurement: (1) Acceleration of train; (2) The speed at which they reached the third sign

Some students took the power train to travel. When the train accelerated evenly on a straight track, one student proposed, "can we use the equipment around us to measure the acceleration of the train?" many students participated in the measurement. The measurement process was as follows: they looked out of the window every 100m   While recording the time with a watch, they observed that the time interval from the first landmark to the second landmark is 5S, and the time interval from the first landmark to the third landmark is 9s. Please find out according to their measurement: (1) Acceleration of train; (2) The speed at which they reached the third sign

(1) Study the process of the train from passing the first road sign to passing the second road sign, with displacement X1 = 100m and time T1 = 5S,
Then the speed at the middle time is V1 = x1
t1
Similarly, study the process of the train from passing the second road sign to passing the third road sign, with displacement x2 = 100m and time T2 = 4S,
Then the speed at the middle time is V2 = x2
t2
Then the acceleration of the train is: a = V2 − v1
t1
2+t2
two
Substitute into the solution: a = 1.11m/s2
(2) According to the kinematic formula, the speed of the train passing the third road sign V3 = V2 + at2
2=27.2m/s
A: (1) the acceleration of the train is 1.11m/s2;
(2) When they reached the third road sign, the speed was 27.2m/s

The car suddenly brakes after driving at the speed of 10m / s for 5 minutes. If the braking process is uniform speed change and the acceleration is 5m / S2, what is the distance the car will travel within 3 seconds after braking?

Vehicle braking time: t0 = 10
5S = 2S, so the car has stopped when braking for 3S, so the displacement within 3S is: x = 10 + 0
two × 2＝10m．
A: the distance traveled by the car within 3 seconds after braking is 10m

A car advances at a uniform speed along a straight highway at a speed of 10m / s. It must brake immediately in case of obstacles. When braking, it makes uniform deceleration with an acceleration of 2m / S2, and the displacement of the car after 6S is () A. 24m B. 25   m C. 26   m D. 30   m

From the meaning of the question, V0 = 10m / s, because of uniform deceleration, a = - 2m / S2, calculate the displacement of the vehicle within 6S
According to the speed time relationship, when the vehicle stops, the speed v = 0, the time t = V − v0
a=0−10
−2s=5s
Therefore, the displacement within 6S of the car is actually the displacement within 5S of the car moving in a straight line with uniform deceleration
So x = v0t + 1
2at2, substituting V0 = 10m / s, a = - 2m / S2 and T = 5S, the displacement x = 25m
Therefore, B

When the train brakes and moves in a straight line with uniform deceleration, after driving for 100 meters in 20 seconds and stopping, what is the speed and acceleration of the train when braking Detailed process

v=100m*2/20s
a=v/20s

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

Initial speed V0 = 5m / s, deceleration a = 2m / S ^ 2 | V1 | = 16m / s ＜ V0, 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 * 13 ^ 2 = - 39m displacement when the speed is zero X1 = V0 ^ 2 / (2a) = 10 ^ 2 / (2 * 2) = 25m, total distance S = 2 | x1 | + | x | = 2 * 25 + 39

When an object moves with uniform deceleration, the initial speed is 10 meters per second, and the acceleration is 1 meter per second, then the average speed of the object within 1 second before stopping the movement A 5.5 meters per second B 5 meters per second C 1 meter per second D 0.5 meters per second

The answer is D. the first second is the last second. You can find 10 stops. Then the first second is 9 seconds later, and the speed is 1 meter per second at 9 seconds, which is the initial speed and the end speed of that second is 0, so the average speed is 0.5