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Car a and car B start from the same place and move in a straight line at a constant speed. The speed of car a is 10m / s, and the speed of car B is 1.5 times that of car A. one minute after car a starts, car B starts to chase car A. how long does it take car B to catch up with car a and how far is it from the starting point?
The two cars have the same distance, s = V A (T + 60s) = v b t, so the time taken by car B is t = v a × 60s v b − V A = v a 60s0.5v a = 60s0.5 = 120s, and the distance from the starting point when catching up is s = v a (T + 60s) = 10m / s × (120s + 60s) = 1800m
The distance between a and B is 100 meters. A is due east of B. at a certain time, a moves due north at a speed of 10 meters per second,
At the same time, B moves due east at the speed of 10m / S. (1) in what direction does B move with a as the reference object? (2) after how long does the distance between a and B last?
(1) B is moving southeast
(2) After five seconds, it's the closest
Let y be the distance after x seconds
y²=(100-10x)²+(10x)²
When x = 5, y is the smallest
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