Question

In an arcade video game, a spot is programmed to move across the screen according to x = 8.09t - 0.686t^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 = 13.0 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? Answer in seconds. (b) At what value of x does this occur? Answer in cms. (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

(a) rest means, instantaneous velocity is zero

And we know that velocity is,

V=dx/dt

Differentiating the given function.

dx/dt = 8.09-2.058t ²

dx/dt=0
t = 1.98 seconds

(b)

Substitute t = 2 in the given function and calculate for the appropriate values of "x"

x=(8.09*1.98)-(0.686*1.98³)

x=10.69 cm

(c)

To determine the acceleration, take the second derivative of the original function, i.e.,

d^x/dt^2 = - 4.116t

and substitute t = 1.98 to determine the acceleration.

=-8.15m/s ²
(d)
Substitute x = 13 intheoriginal function, i.e.,

15 = 8.09t - 0.686t³

t=4.67 seconds