In an arcade video game, a spot is programmed to move across the screen accordin

Question

In an arcade video game, a spot is programmed to move across the screen according to x = 7.50t - 0.850t3, where x is distance in centimeters measured from the left edge of the screen and t is time in seconds. When the spot reaches a screen edge, at either x = 0 or x = 18.0 cm, t is reset to 0 and the spot starts moving again according to x(t).


When does it first reach an edge of the screen after t = 0?

Explanation / Answer

Hey Barca, hope this helps! a) The spot is at rest when velocity is 0. You have the position equation. Take the first derivative to get the velocity equation: dx/dt = 7.5 - 2.55t2 Set this equal to 0 to find when the spot is at rest: 7.5 - 2.55t2 = 0 7.5 = 2.55t2 t2 = 2.94 t = 1.7 b) Plug t = 1.7 to the position equation to find x x = 7.5(1.7) - .85(1.7)^3 x = 12.75 - 4.18 x = 8.57cm c) You have t now, but now you need accel. You need to take the derivative of the velocity equation now: dv/dt = -5.1t Plug in t and you get -5.1(1.7) = -8.67cm/s^2 f) you have the position equation, if you graph it, you can see that x increases, but never reaches 18 (the right edge), instead it turns around (at t = 1.7) and goes back to the starting edge. Thus plug in 0 to the position equation (when it reaches the left edge again): 0 = 7.50t - 0.850t3 (obviously t=0 is a solution, but we want a t > 1.7) 0 = t(7.5 - .85t2) 0 = 7.5 - .85t2 0 = (2.74 - .92t)(2.74 + .92t) 0 = 2.74 - .92t .92t = 2.74 t = 2.98

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