Question

A Canadian Goose and Great Blue Heron took off in opposite directions . The goose flew at 20 mph , and the heron flew 10 mph . When they landed, they were 180 miles apart .Altogether , they flew for 14 hours . How long did each bird fly?

Answer 1

45 mph because the goose has longer wingspan

Answer 2

Answer:

The canadian Goose fly 4 hours and the Great Blue Heron fly 10 Hours

Step-by-step explanation:

Lets call, $v_A, d_A$ and $t_A$ the velocity, distance and time of fly for the Canadian Goose respectively and  $v_B, d_B$ and $t_B$ the velocity, distance and time of fly for the Great Blue Heron respectively.

Additionally, it is necessary to know that:

$d_A=v_A*t_A\\d_B=v_B*t_B$

So, from the question, we know that $v_A$ is equal to 20 mph and $v_B$ is equal to 10 mph.

Them, from the sentence: when they landed, they were 180 miles apart, we can formulate the equation 1 as:

$d_A+d_B=180\\(v_A*t_A) + (v_B*t_B)=180\\(20*t_A)+(10*t_B)=180$

And from the sentence: Altogether, they flew for 14 hours, we can formulate the equation 2 as:

$t_A+t_B=14\\t_B=14-t_A$

So, replacing $t_B$ from equation 2 on equation 1 and solving for $t_A$, we get:

$20t_A+10t_B=180\\20t_A + 10(14-t_A)=180\\20t_A + 140-10t_A=180\\10t_A + 140=180\\10t_A=180-140\\10t_A=40\\t_A=\frac{40}{10} \\t_A=4$

Replacing $t_A$ on equation 2, we get:

$t_B=14-t_A\\t_B=14-4\\t_B=10$

Finally, $t_A$ is equal to 4 hours and $t_B$ is equal to 10 hours. It means that the canadian Goose fly 4 hours and the Great Blue Heron fly 10 Hours.