In an arcade video game, a spot is programmed to move across the screen according to x = 9.00t - 0.750t³, 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 = 15.0 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 (including sign) when this occurs? (d) Is it moving right or left just prior to coming to rest? (e) Just after? (f) At what time t > 0 does it first reach an edge of the screen?

a) The spot is at rest at t = 2.00 s.

b) The spot is at rest at x = 12 cm.

c) The acceleration at t = 2.00 s is - 9.00 cm/s².

d) Before coming to rest, the spot is moving to the right.

e) After t = 2.00, the spot is moving to the left.

f) The spot reaches the edge of the screen at t = 3.46 s.

Explanation:

Hi there!

a) The velocity is defined as the variation of the position over time. If the time interval is very small and being "x" the position of the spot and "t" the time, the instantaneous velocity can be expressed as follows:

v = dx/dt

The position function is the following:

x (t) = 9.00 · t - 0.750 · t³

dx/dt is equal to the derivative of x (t):

v = dx/dt = 9.00 - 3 · 0.750 · t²

v = 9.00 - 2.25 · t²

When the spot is at rest, v = 0.

0 = 9.00 - 2.25 · t²

Solving for t:

-9.00 / - 2.25 = t²

t = 2.00 s

The spot is at rest at t = 2.00 s.

b) To find the value of the position at t = 2.00, we have to evaluate the function x (t) at t = 2.00 s.

x (2.00) = 9.00 · (2.00) - 0.750 · (2.00) ³

x (2.00) = 12 cm

The spot is at rest at x = 12 cm

c) The acceleration is defined as the variation of the velocity over time. If the time is very small, we get the instantaneous acceleration:

a = dv/dt

Then, we have to derivate the velocity function:

v (t) = 9.00 - 2.25 · t²

a = dv/dt = - 2 · 2.25 t

a (t) = - 4.50 · t

At t = 2.00 s, the acceleration will be a (2.00):

a (2.00) = - 4.50 · 2.00

a (2.00) = - 9.00 cm/s²

The acceleration at t = 2.00 s is - 9.00 cm/s²

d and e) The velocity indicates the direction of displacement. If it is positive, the spot is moving to the right (away from the origin at x = 0) and if it is negative the spot is returning to the origin, moving to the left. Let's evaluate the velocity function at t = 1.99 and 2.01:

v (1.99) = 9.00 - 2.25 · (1.99) ²

v (1.99) = 0.09 cm/s

v (2.01) = 9.00 - 2.25 · (2.01) ²

v (2.01) = - 0.09 cm/s

Before coming to rest, the spot is moving to the right. After t = 2.00, the spot is moving to the left.

f) We have to find the time at which x = 0. Notice that since the acceleration is always negative and the spot comes to rest at x = 12 cm and then returns to the origin, the spot will never reach the edge at x = 15 cm. So, let's find the time at which x = 0:

x = 9.00 · t - 0.750 · t³

At x = 0:

0 = 9.00 · t - 0.750 · t³

Solving for t:

0 = t (9.00 - 0.750 t²)

0 = 9.00 - 0.750 t²

-9.00 / - 0.750 = t²

t = 3.46 s

The spot reaches the edge of the screen at t = 3.46 s.