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 e quation models the distance y the spring stretches with weight x attached to it?
1 answer:
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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