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Car a and car B are moving in the same direction on the same straight line. Car a is driving at a constant speed of 10m / s, and car B is moving at a static speed of 2m / S2 at 10m behind the home 1. How long does it take for car B to catch up with car a? What are the speeds of car a and car B at this time? 2. How long does it take to catch up with Qianjing? What is the farthest distance between them
(1) T ^ 2-10 = 10t, we get t = (35 under the root) - 5, because a is uniform, then the velocity is 10 meters per second, B is v = at, then v = (35 under the double root) - 10
(2) When the velocity of a and B is equal, the distance is the farthest, then t = V / A, t = 10 / 2 = 5S, then the displacement of a is 50, and the displacement of B is s = 1 / 2at ^ 2 = 25, so the distance is 50-25 + 10 = 35
Car a and car B start from the same place at the same time and move in the same direction, in which car a runs at a uniform speed of 10m / S2 and car B drives from the same place at an acceleration of 2m / S2
Static start,
1) How long does it take for car B to catch up with car a? What is the relationship between the speed of car a and car B?
2) How long does it take to catch up with Qianjing? What is the relationship between the speed of the two at this time?
1. Let's catch up after x seconds, then the displacement of a and B in x seconds should be the same, so we get the equation: half times the acceleration of B times the square of X (displacement formula) = the velocity of a times X. then we get x = 10 or 0 by solving X. because x = 0 does not conform to the meaning of the problem, so x = 10s
2. Because B's speed is slower than a at the beginning, it's equivalent to using B as a reference. A is moving forward while B is not moving, so the distance between them keeps increasing. When they are at the same speed, they are relatively stationary. Then B starts to chase a upward. Because B's speed is still increasing, they are farthest when they are at the same speed
First, find out how long the speed of a and B is the same. You should know, 10 / 2 = 5 seconds. After 5 seconds, the speed of a and B is equal. In these 5 seconds, a walks 5 times 10 = 50 meters, B walks half times B's acceleration times 25 = 25 meters, so the maximum distance is 25 meters
Car a and car B are moving in a straight line at a constant speed. Car a is 20 meters in front of car B and moves two seconds earlier than car B. the speed of car a is 2m / s,
The speed of car B is 10.8km/h
1. How long does it take car B to catch up with car a?
2. When car B overtakes car a, how far is the home car from
(1) The speed of car B is 10.8x1000m/3600s = 3m / s
Let car B take T seconds to catch up with car a, then car a takes 2 seconds less, that is, it takes (T-2) seconds
3xt = (T-2) x2 + 20, the solution is t = 16
(2) It can be seen from (1) that it took 16 seconds for car B to catch up with car a, and it took 14 seconds for car a to leave the starting place
The distance is 14x2m = 28m
The two trains run in opposite directions. The speed of car a is 54km / h, and that of car B is 46km / h. when they are 5km / h apart, the speed of car a is 54km / h
54Km/h=54000m/3600s=15M/S
46Km/h=46000m/3600s=115/9M/S
1. Time for car B to hear flute: 5000 / (115 / 9 + 340) = 5000 / 3175 / 9 = 1800 / 127S
2. At this time, the distance between two cars is 5000 - (15 + 115 / 9) * 1800 / 127 = 5000 * 117 / 127m
3. Echo of car a: 1800 / 127 + (5000 * 117 / 127) / (15 + 340)
It seems that the speed of car B is 36km / h, which is easier to calculate
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