Chapter 2, Problem 71 In an arcade video game, a spot is programmed to move acro

Question

Chapter 2, Problem 71

In an arcade video game, a spot is programmed to move across the screen according to x = 7.00t - 0.575t3, where x is the distance in centimeters measured from the left edge of the screen and t is time in seconds. When the spot reaches a screen edge, either at x = 0 or x = 11.0 cm, t is reset to 0 and the spot starts moving again according to x(t). (a) At what time after starting is the spot instantaneously at rest? (b) At what value of x (in cm) does this occur? (c) What is the spot's acceleration (in cm/s^2, including sign) when this occurs? (d) At what time t > 0 does the spot first reach an edge of the screen?

Explanation / Answer

a) The spot's instantaneous velocity is x'(t) = dx/dt = 7.0 - 1.725t². When the spot is instantaneously at rest, x'(t) = 0, which gives: x'(t) = 7 - 1.725t² = 0 t² = 7/1.725=4.06 t = 2.014 s b) The position at that time is: x(2.014) = 7(2.014) - 1.725(2.014)³ = 9.25*10^-3 c) The instantaneous acceleration is x"(t) = d²x/dt² = -3.45t. At t = 2.0143.489, the acceleration is -3.45(2.014) = -6.948 d) the spot moves across the screen to a position equal to x = 16.2668, and then reverses direction and moves back toward the left until it reaches the left edge of the screen. Setting the position, x(t), equal to zero and solving for t gives: x(t) = 7t - 0.575t³ = 0 Factoring x(t) gives: t(7 - 0.575t²) = 0 One solution is t = 0. The other solution is found from: 7 - 0.575t² = 0 t² = 7 / 0.575 t² = 12.17 t = 3.489

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