The position of a particle moving along the x axis depends on the time according

Question

The position of a particle moving along the x axis depends on the time according to the equation x = ct^2 - bt^5, where x is in meters and t in seconds. Let c and b have numerical values 2.5 m/s^2 and 1.6 m/s^5, respectively. From t = 0.0 s to t = 1.3 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s. Note: Please put answers in a) b) c) form ... If possible show all the work you can. Or -- for an easier 5 star -- tell me the equation to find this and which variables to use for 1 and I will figure out the rest.

Explanation / Answer

x(0) = c*0 - b*0 = 0. x(4) = c*4^3 - b*4^2 = 3.6*64 - 1.0*16 = 214.4 You don't need a derivative. You are asked for the difference between two positions. x is the position. Edit: You're right, I messed up the formula. x(4) = 3.6*4^2 - 1.0*4^3 = -6.4. The displacement is -6.4 m. Edit again: Aaaaah. I see what it's asking. The distance it covers is the total distance traveled. So we do need to check the velocity after all. v = dx/dt = 2ct - 3bt^2. First question: Where is this velocity 0, which is where the particle turns around? Answer: t(2c - 3bt) = 0 when t = 0 and when t = 2c/3b = 7.2/3 = 2.4. So it moves in the +x direction up till t = 2.4, and in the -x direction after that. At t = 2.4 the position is 3.6*2.4^2 - 1.0*2.4^3 = 6.9 m. It has traveled 6.9 meters so far. At t = 4 we know the position is x = -6.4. It has traveled an additional (6.9 - (-6.4)) = 13.3 meters to get there. Total distance traveled, back and forth = 6.9 + 13.3 = 20.2 m.

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