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 = 11.5t - 0.598t3, where x is the distance in centimeter 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 = 23.0 cm, t is reset to 0 and the spot starts moving again according to x(t). At what time after is the spot instantaneously at rest? At what value of x (in cm) does this occur What is the spot's acceleration (in cm / s ^ 2, including sign) when this occurs? At what time t > 0 does the spot first reach an edge of the screen?

Explanation / Answer

a)

dx/dt=11.5-1.794t^2=0

t^2=6.410

t=2.53 s

b)

x=11.5*2.53-0.598*2.53^3

=19.41 m

c)

dv/dt=-3.588*2.53

         =-9.077 m/s^2

d)

you have the position equation, if you graph it, you can see that x increases, but never reaches 23 (the right edge), instead it turns around (at t = 2.53) and goes back to the starting edge. Thus plug in 0 to the position equation (when it reaches the left edge again):

0 = 11.5t - 0.598t^3 (obviously t=0 is a solution, but we want a t > 2.53)

11.5-0.598t^2=0

t^2=19.23

t=4.38 s

PLS RATE IT

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