Question

A brick becomes dislodged from the top of a building (at a height of 1280 feet) and falls to the sidewalk below. (Let the position function for a free-falling object be
−16t2 + v0t + h0.)
(a) Write the position, velocity, and acceleration functions of the brick

Answer 1

As the accelerations is constant therefore the position of brick is

x(t) = at^2 / 2+ xo = -16t^2 + 1250

Step-by-step explanation:

• The rate of change of velocity is given by the equation below.
• V(t)=a t +Vo
• As we know that the initial velocity of the brick is zero. Thereofre, Vo = zero
• Now V(t) = a(t)
• The unit of acceleration is ft / s^2 and the unit of velocity is ft/ s^2.
• Thus the unit of at becomes ft.s / s^2 = ft/s
• As the accelerations is constant therefore the position of brick is
• x(t) = at^2 / 2+ xo = -16t^2 + 1250

Answer 2

Answer:

As the accelerations is constant therefore the position of brick is

x(t) = at^2 / 2+ xo = -16t^2 + 1250

Step-by-step explanation:

The rate of change of velocity is given by the equation below.

V(t)=a t +Vo

As we know that the initial velocity of the brick is zero. Thereofre, Vo = zero

Now V(t) = a(t)

The unit of acceleration is ft / s^2 and the unit of velocity is ft/ s^2.

Thus the unit of at becomes ft.s / s^2 = ft/s

As the accelerations is constant therefore the position of brick is

x(t) = at^2 / 2+ xo = -16t^2 + 1250

Step-by-step explanation:

Hope this answer will help you.✌️