Question

In an arcode video game a spot is programmed to move across the screen to x=9.00t-0.750t^3, where x is the distance in cm and t is the time in seconds. when the spot reaches a screen edge, at either x=0 or x=15.0cm. t is reset to 0 and spot start moving again according to x(t)
a) at what time after starting is the spot instantaneously at rest.
b) where does it occur
c) what is its acceleration when it occurs
d)in what direction is it moving just prior to coming to rest.
dii)just after
e) when does it first reach an edge of the seven after t=0.

Answer 1

Answer on Question #66130-Physics-Mechanics-Relativity

In an arcade video game a spot is programmed to move across the screen to $x=9.00t-0.750t^3$, where $x$ is the distance in cm and $t$ is the time in seconds. When the spot reaches a screen edge, at either $x=0$ or $x=15.0\mathrm{cm}$, $t$ is reset to 0 and spot start moving again according to $x(t)$

a) at what time after starting is the spot instantaneously at rest.

b) where does it occur

c) what is its acceleration when it occurs

d)ii)in what direction is it moving just prior to coming to rest.

d)ii)just after

e) when does it first reach an edge of the seven after $t = 0$.

Solution

a)

$$v = \frac{dx}{dt} = 9.00 - 3 \cdot 0.750t^2 = 0$$ $$t = 2\,s.$$

b)

$$x(2) = 9.00(2) - 0.750(2)^3 = 12\, \text{cm}.$$

c)

$$\frac{dv}{dt} = -3 \cdot 2 \cdot 0.750t.$$ $$\frac{dv}{dt}(2) = -3 \cdot 2 \cdot 0.750(2) = -9\,\frac{\text{cm}}{\text{s}^2}.$$

d)i) Positive.

ii) Negative.

e)

$$x = 0 = 9.00t - 0.750t^3$$ $$t = 3.46\,s.$$

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