Passenger A on the bus gets off after arriving at station A, and walks in the opposite direction of the bus. After 40 seconds, the bus arrives at station B, and B immediately gets off and chases A. If B’s speed is twice as fast as A’s, is one-fifth of the speed of the bus, how many seconds does it take for B to catch up with A after getting off the bus?
公共汽车上的乘客甲到达A站后下车，向公共汽车行驶方向的反方向行走，40秒后，公共汽车到达B站，乙立即下车追赶甲，如果乙的速度比甲快一倍，是公共汽车速度的五分之一，那么乙下车后经多少秒能追上甲．
<gadget id="calculator">1 * 40</gadget>
<output>40</output>

<gadget id="calculator">1 + 1</gadget>
<output>2</output>

<gadget id="calculator">1 / 5</gadget>
<output>1/5 = around 0.2</output>

<gadget id="calculator">2 / (1/5)</gadget>
<output>10</output>

<gadget id="calculator">10 * 40</gadget>
<output>400</output>

<gadget id="calculator">40 + 400</gadget>
<output>440</output>

<gadget id="calculator">440 / 1</gadget>
<output>440</output>

<result>440</result>