# Chapter 2 - Motion Along a Straight Line - Problems: 5a

x = 0

#### Work Step by Step

The position is given by the equation $x = 3t - 4t^2 + t^3$ where x is in meters and t is in seconds To find the position at t = 1 s we need to find x(1), or in other words, plug in 1 for t in the equation. This gives us: $x = 3(1) - 4(1)^2 + (1)^3 = 3 - 4(1) + 1 = 3-4+1 = 0$ Therefore at t =1, the position of the object is x = 0

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