For a train moving in a straight line with uniform speed change, the speed increases from 10m / s to 15m / s within 50s. What is the acceleration of the train?

For a train moving in a straight line with uniform speed change, the speed increases from 10m / s to 15m / s within 50s. What is the acceleration of the train?

Acceleration a = (VT VO) / T = (15-10) / 50 = 0.1m/s ^ 2

The speed of the train is 8m / s. when the engine is turned off and the train moves forward for 70m, the speed will be reduced to 6m / s. if it passes another 50s, what is the further distance of the train

From V22 − V12 = 2aX, a = V22 − V12
2x＝36−64
140＝−0.2m/s2．
Time required for train to stop from 6m / s t0 = 6
0.2s＝30s＜50s．
Therefore, the displacement within 50s is equal to that within 30s
Then x '= 0 − V22
2a＝−36
−0.4m＝90m．
A: the distance of the train is 90m

The speed of the train is 8m / s. when the engine is turned off and the train moves forward for 70m, the speed will be reduced to 6m / s. after another 50s, the distance of the train will be (the train will slow down evenly after turning off the engine) () A. 50   m B. 90   m C. 120   m D. 160   m

By V
two
two
−v
two
one
= 2aX: a = v
two
two
−v
two
one
2x=62−82
two × 70=-0.2m/s2
Time required for train to stop from 6m / s: t0 = 6
0.2=30s
Therefore, the displacement within 50s is equal to that within 30s
Then x '= 0 − V
two
two
2a=−36
−0.4=90m
A: the distance of the train is 90m

For a train moving in a straight line with uniform speed change, the speed increases from 10m / s to 15m / s within 50s. What is the acceleration of the train

V=Vo+at
15=10+a*50
a=0.1

The speed of the train is 8m / s. when the engine is turned off and the train moves forward for 70m, the speed will be reduced to 6m / s. if it passes another 50s, what is the further distance of the train

From V22 − V12 = 2aX, a = V22 − V12
2x＝36−64
140＝−0.2m/s2．
Time required for train to stop from 6m / s t0 = 6
0.2s＝30s＜50s．
Therefore, the displacement within 50s is equal to that within 30s
Then x '= 0 − V22
2a＝−36
−0.4m＝90m．
A: the distance of the train is 90m

A train runs at a constant speed on a flat track at the speed of 20m / s, and the resistance of the train is 9 * 10 ^ 4N. Then what is the traction generated by the engine? One Minute... A train runs at a constant speed on a flat track at the speed of 20m / s, and the resistance of the train is 9 * 10 ^ 4N. Then what is the traction generated by the engine? How much work does traction do in a minute?

1. Because it is a uniform speed, it means that the force on the vehicle is balanced, so the traction force is equal to the resistance
F = 90000n
2. Work done by traction force in one minute: w = FS = FVT = = 90000 * 20 * 60 = 108000000j = 1.08 * 10 ^ 8j