An observer on the platform stood at the front of the first carriage as the train started There is an observer on the platform. When the train starts, he stands near the front end of the first carriage. The first carriage passes him in 10 seconds. Suppose the train moves in a straight line with uniform acceleration, what is the time interval for the seventh carriage to pass him? If there are 10 carriages, what is the time interval for all the carriages to pass him/

An observer on the platform stood at the front of the first carriage as the train started There is an observer on the platform. When the train starts, he stands near the front end of the first carriage. The first carriage passes him in 10 seconds. Suppose the train moves in a straight line with uniform acceleration, what is the time interval for the seventh carriage to pass him? If there are 10 carriages, what is the time interval for all the carriages to pass him/

S = v0t + at2 according to this formula, it is assumed that the length of each carriage is s, and the past time of the first carriage is 10 seconds
For the uniformly accelerated linear motion with zero initial velocity, the following law holds
S = at2, a = s / T2, then the passing time of all seven cars is under the root sign (7S / a)
The driving time of the first six sections is under the root sign (6S / a). After calculation, it can be seen that the time interval of the seventh section is 10 (under the number 7) - under the root sign 6) = 2 seconds, because in the first section, you said 10 seconds, so the specific time is smaller than 2 seconds
The total time of 10 cars is 10 times 10 under the root sign, which is about 31.622 seconds

The train accelerates on the flat track with an acceleration of 0.98m/s2. A passenger in the carriage reaches out of the window and releases an object from the place 2.5m above the ground. Regardless of the air resistance, the horizontal distance between the object and the passenger is ()
A. 0b. 0.25mc. 0.50md

Let v be the speed of the train when the ball is released. According to h = 12gt2, the time of horizontal throwing motion of the object is t = 2hg = 2 × 2.510 = 22s. The horizontal displacement of the object is X1 = vt. the displacement of the train is x2 = VT + 12at2, then the displacement difference △ x = x2-x1 = 12at2 = 12 × 0.98 × 12S ≈ 0.25m

When the train runs on the horizontal track at the acceleration of 1m / S2, a passenger reaches out of the window and releases an object freely from 2.5m above the ground. Regardless of the air resistance, the horizontal distance between the object and the passenger is ()
A. 0mB. 0.5mC. 0.25mD. 1m

Let the speed of the train be V & nbsp; & nbsp; & nbsp; & nbsp; when the object is just released, the time of horizontal throwing motion of the object is t = 2H / G & nbsp; & nbsp; & nbsp; & nbsp; the displacement of uniform acceleration motion of the train x1 = VT + 12at2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; the horizontal displacement of horizontal throwing motion x2 = vt & nbsp; & nbsp; & nbsp; & nbsp; so the distance between the object and the passengers is s = x1-x2 = 12at2 = AHG = 1 × 2.510m = 0.25m, so C is selected