Question

In an arcade video game, a spot is programmed to move across the screen according to x = 8.79t - 0.658t^3, 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 = 15.0 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 does this occur? (c) What is the spot's acceleration when this occurs? (d) At what time t > 0 does the spot first reach an edge of the screen?

Answer 1

Answer on Question #51375, Physics, Mechanics Kinematics Dynamics

In an arcade video game, a spot is programmed to move across the screen according to $x = 8.79t - 0.658t^3$ , 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 $x = 15.0cm$ , $t$ 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 $\mathbf{x}$ does this occur?

(c) What is the spot's acceleration when this occurs?

(d) At what time $t > 0$ does the spot first reach an edge of the screen?

Solution:

(A) The spot is instantaneously at rest if $x = 0$ or $x = 15.0cm$ . Then if $x = 0$ $8.79t - 0.658t^3 = 0 \Rightarrow t(8.79 - 0.658t^2) = 0$

$$t _ {1} = 0 s$$ $$t _ {2, 3} = \pm \sqrt {\frac {8 . 7 9}{0 . 6 5 8}} = \pm 3. 6 5 s$$

We consider only physically correct solutions $(t > 0)$ .

Fig.1

If $x = 15.0cm$ than $8.79t - 0.658t^3 = 15$

We built the dependence of $x(t)$ using mathematical software (see Fig.1). From Fig.1 it is clear that $x$ never gets 15cm.

(B) From Fig. 1 it clear that $x \in [0, x_{\max}]$ . So $\frac{dx}{dt} = 8.79 - 3 \cdot 0.658t^2 = 0 \Rightarrow t = 2.11$ , then

$$\frac {d ^ {2} x}{d t ^ {2}} = - 6 \cdot 0. 6 5 8 t \Rightarrow \frac {d ^ {2} x}{d t ^ {2}} (2. 1 1) = - 6 \cdot 0. 6 5 8 \cdot 2. 1 1 = - 8. 3 3 < 0 \Rightarrow t _ {\max } = 2. 1 1.$$ $$x _ {\max } (2. 1 1) = 8. 7 9 \cdot 2. 1 1 - 0. 6 5 8 \cdot 2. 1 1 ^ {3} = 1 2. 3 7 c m$$ $$x \in [ 0, 1 2. 3 7 ]$$

(C) The spot's acceleration is $a(t) = \frac{d^2x}{dt^2} = -6\cdot 0.658t$

$$a (0) = 0$$ $$a (3. 6 5) = - 6 \cdot 0. 6 5 8 \cdot 3. 6 5 = - 1 4. 4 1 m / s$$

(D) The spot is never reach an edge of the screen (see Fig.1)