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A spring stretches to 22 cm with a 70 g weight attached to the end. With a 105 g weight attached, it stretches to 27 cm. Which equation models the distance y the spring stretches with weight x attached to it?

Respuesta :

Assume that the spring behaves linearly so that
y = a + bx
where 
x = weight of the mass
y =  stretch of the spring
a,b are constants

When x = 70 g, y = 22 cm, therefore
a + 70b = 22             (1)

When x = 105g,  y = 27 cm, therefore
a + 105b = 27           (2)

Subtract (1) from (2).
a + 105b - (a + 70b) = 27 - 22
35b = 5
b = 0.1429

From (1), obtain
a = 22 - 70*0.1429 = 12

Answer:
The required equation is
y = 12 + 0.1429x

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