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The hare begins his morning jog an unknown distance from his burrow. He runs at a constant rate, in a straight line away from his burrow. After jogging for 35 seconds, he passes a boulder he know is 220 feet in front of his home. After two and a half minutes he passes a tree that he knows is 680 feet from his home.

1.) Create an equation
2.) Explain what the terms, coefficients, and factors or your equation mean in the context of his problem

Respuesta :

After jogging for 35 seconds, he passes a boulder he know is 220 feet in front of his home. After two and a half minutes he passes a tree that he knows is 680 feet from his home.

Let x be the time taken in seconds

Let y be the distance from the home

After jogging for 35 seconds, he passes a boulder he know is 220 feet in front of the home. so x is 35 seconds and y is 220. (35,220)

After two and a half minutes he passes a tree that he knows is 680 feet from his home. so x is 150 second and y is 680. (150, 680)

Hare runs at constant rate so equation is y= mx + b

To frame the equation we need to find out slope 'm' and 'b'

(35,220) and (150, 680)

Slope m = [tex] \frac{y_2-y_1}{x_2 - x_1} [/tex]

m = [tex] \frac{680-220}{150 - 35} [/tex]

m= 4 feet per second

so y= 4x +b

Now we find out b using (35,220)

220= 4(35) +b

80 = b

The required equation is y= 4x + 80

x is the time taken in seconds

y is the distance from the home in feet

4 represent speed of the hare.

Hare started jogging 80 feet from the home.


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