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 the relation x(t) = 9.00t-0.75t 3, 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 = 15 cm, t is reset to 0 and the spot starts moving again according to x(t) a. b. c. d. e. f. At what time after starting is the spot instantaneously at rest? At what value of x does this occur? What is the spot's acceleration (including sign) when this occurs? Is it moving right or left just prior to coming to rest? Just after? At what time t >0 does it first reach an edge of the screen?

Explanation / Answer

a)

x(t) = 9 t - 0.75 t3

taking derivative both side relative to "t"

v(t) = dx(t)/dt = 9 - 2.25 t2

at rest , v(t) = 0

9 - 2.25 t2 = 0

t = 2 sec

b)

at t = 2

x(2) = 9 (2) - (0.75) (2)3 = 12 cm

c)

v(t) = 9 - 2.25 t2

acceleration is given as

a(t) = dv(t)/dt = 0 - 4.5 t

at t = 2

a(2) = - 4.5 (2) = - 9 m/s2

d)

it is moving towards right

e)

just after it is moving towards left

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