Unformatted text preview: k = 10, so s ( t ) = 3 t 2 / 2 + 10. For example, at t = 1 the object is at position 3 / 2 + 10 = 11 . 5. This is certainly the easiest way to deal with this problem. Not all similar problems are so easy, as we will see; the second approach to the problem is more diﬃcult but also more general. We start by considering how we might approximate a solution. We know that at t = 0 the object is at position 10. How might we approximate its position at, say, t = 1? We know that the speed of the object at time t = 0 is 0; if its speed were constant then in the ﬁrst second the object would not move and its position would still be 10 when t = 1. In fact, the object will not be too far from 10 at t = 1, but certainly we can do better. Let’s 139...
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 Spring '07
 JonathanRogawski
 Math, Calculus, Derivative, Position

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