Simple Physics Problem: Finding c in a Particle's Position Equation at t = 3.0 s

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SUMMARY

The discussion centers on calculating the constant c in the position equation of a particle, given by x = 3.0 m + (4.0 m/s)t + ct² - (2.2 m/s³)t³. At t = 3.0 s, the force acting on the 1.2 kg particle is 36 N in the negative direction. By applying Newton's second law, F = MA, and differentiating the position equation twice to find acceleration, the equation 36 = (1.2)(2c - 39.6) is established. Solving this yields the value of c.

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Homework Statement



A 1.2 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3.0 m + (4.0 m/s)t + ct2 - (2.2 m/s3)t3, with x in meters and t in seconds. The factor c is a constant. At t = 3.0 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c?

Homework Equations





The Attempt at a Solution




I double dif. to get a = 2c-13.2t and plug in t = 3 => 2c-39.6 right? F=MA so its 36 = (1.2)(2c-39.6)?
 
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