Question

005 (part 1 of 2) 10.0 points

Answer 1

A constant speed motion is one in which equal distances are covered in equal times

The bird travels a cumulative distance of 7.$\overline 3$ km

The reason the above value is correct is as follows:

The known parameters are;

Speed of the runner, $v_r$ = 2.1 km/hr

The location where the bird begins to fly to the finish line = When the runner is 4.4 km from the finish line

The speed of the bird, $v_b$ = 10.5 km/hr = 5 times the runners speed

The bird reaches the finish line, turns, and returns back to the runner

Required:

The find cumulative distance traveled by the bird

Solution:

The distance the bird travels is five times the distance the runner travels, therefore,

Let x represent the distance the runner ravels before the bird returns, we have;

The distance the bird travels = 4.4 + 4.4 - x = 8.8 - x

The distance the runner travels = x

The time the runner runs x km = The time the bird flies (8.8 - 4) km

$From \ velocity = \dfrac{Distance }{Time}$, we have;

$Time= \dfrac{Distance }{Velocity}$

Given the time taken by the runner is equal to the time taken by bird, while running, we have;

$Time= \dfrac{x}{2.1} = \dfrac{8.8 - x}{10.5}$

Therefore;

10.5·x = 2.1·(8.8 - x) = 2.1×8.8 - 2.1·x

10.5·x + 2.1·x =  2.1×8.8 = 18.84

12.6·x = 18.84

$x = \dfrac{18.84}{12.6} =\dfrac{22}{15} = 1.4 \overline 6$

$The \ distance \ the \ bird \ travels = 8.8 - \dfrac{22}{15} \approx \dfrac{22}{3} = 7. \overline 3$

The cumulative distance the bird travels is 7.$\overline 3$ km

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