Two ships are 10 km apart and moving toward each other at 6.0 km/h and 4.0 km/h, respectively. a bird flies continuously back and forth between them at 20 km/h. what is the total distance traveled by the bird when the ships meet?

20 km

Step-by-step explanation: Let's assume a trajectory of Ships A (which is moving at 6 km/h) and B (which is moving at 4 km/h)

A            10 km           B  

And let's assume that who is going to the right is the positive.    

Space (S) = S₀ + vt

For A, its space is SA = 0 + 6t

For B, its space is SB = 10 - 4t

They meet at SA = SB;

6t = 10 - 4t

6t + 4t = 10

10t = 10

t = 1h

If the bird is at 20 km/h and it will fly for 1h, it will travel 20 km

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.