A brick becomes dislodged from the top of a state building (at a height of 1290 feet) and falls to the sidewalk below.
(a) Write the position, velocity, and acceleration functions of the brick.
s(t) =
v(t) =
a(t) =
(b) How long does it take the brick to hit the sidewalk?
(c) How fast is the brick traveling when it hits the sidewalk?
(a)
s = 1290 - 16t^2
... for v,a
(b) solve for t:
1290 - 16t^2 = 0
(c) plug that t into v(t)
(a) Alright, let's get mathematical, but with a twist!
First, we need to set up our equations. Let's assume that the height of the state building is represented by s(t), the velocity is represented by v(t), and the acceleration is represented by a(t).
s(t) = -16t^2 + 1290
v(t) = -32t
a(t) = -32
Why all the negatives, you ask? Well, gravity has a habit of bringing things down, so we're keeping it real here. And by real, I mean negative.
(b) Now, to figure out when the brick hits the sidewalk, we need to find the value of t when s(t) equals 0.
0 = -16t^2 + 1290
Solving for t, we get:
16t^2 = 1290
t^2 = 80.625
t ≈ ±8.98
But since time cannot be negative here, we'll only consider the positive value:
t ≈ 8.98 seconds
(c) Finally, to find out how fast the brick is traveling when it hits the sidewalk, we need to find v(t) at t = 8.98.
v(8.98) = -32(8.98)
v(8.98) ≈ -287.36 feet per second
So, the brick hits the sidewalk with a speed of approximately 287.36 feet per second. Oh boy, that's quite a drop!
(a) To find the position, velocity, and acceleration functions of the brick, we can use the standard equations of motion under constant acceleration.
Let's assume the initial position of the brick is at s(0) = 0 feet.
(i) Position function:
The position function can be determined using the equation:
s(t) = s(0) + v(0)t + (1/2)at^2, where s(t) is the position at time t, v(0) is the initial velocity, a is the constant acceleration, and t is the time.
For the present scenario, the initial velocity v(0) is 0 ft/s, and the constant acceleration a is due to gravity, approximately -32.2 ft/s^2 (considering downward direction as negative). Therefore, the position function becomes:
s(t) = 0 + 0t - (1/2)(32.2)t^2 = -16.1t^2
(ii) Velocity function:
The velocity function can be found by differentiating the position function with respect to time (t):
v(t) = ds(t)/dt
Differentiating s(t) = -16.1t^2:
v(t) = d/dt(-16.1t^2) = -32.2t
(iii) Acceleration function:
The acceleration function remains constant at -32.2 ft/s^2.
Therefore, the functions for position, velocity, and acceleration of the brick are:
s(t) = -16.1t^2
v(t) = -32.2t
a(t) = -32.2
(b) To find the time it takes for the brick to hit the sidewalk, we need to determine when the position function s(t) becomes equal to the height of the sidewalk, s(t) = 0 feet.
Setting s(t) = 0 and solving for t:
-16.1t^2 = 0
t^2 = 0
t = 0
Since time cannot be negative, the value of t = 0 is not considered.
Therefore, the brick hits the sidewalk at t = 0 seconds.
(c) To find the speed of the brick when it hits the sidewalk, we can substitute t = 0 into the velocity function v(t):
v(0) = -32.2(0) = 0 ft/s
Hence, the brick is traveling at 0 ft/s when it hits the sidewalk.
(a) To determine the position, velocity, and acceleration functions of the brick, we need to use the equations of motion under constant acceleration.
Let's assume that the initial position of the brick is 0 ft, and the positive direction is upwards.
1. Position Function (s(t)):
The position function describes the vertical displacement of the brick at any given time (t). For this problem, we can use the equation:
s(t) = s₀ + v₀t + (1/2)at²
Since the initial position (s₀) is 0 ft and the initial velocity (v₀) is 0 ft/s (as the brick initially is at rest), we can simplify the equation to:
s(t) = (1/2)at²
2. Velocity Function (v(t)):
The velocity function describes the rate at which the position changes with respect to time. It is the derivative of the position function. Taking the derivative of the position function, we have:
v(t) = ds(t)/dt
Since the derivative of a constant (1/2a) is zero, the velocity function simplifies to:
v(t) = at
3. Acceleration Function (a(t)):
The acceleration function represents the rate at which the velocity changes with respect to time. In this case, the acceleration is constant, as it is due to the force of gravity. Hence, the acceleration function is a constant value and can be written as:
a(t) = -g
where g is the acceleration due to gravity, approximately 32.2 ft/s².
(b) To find how long it takes the brick to hit the sidewalk (t_hit), we can set the position function equal to the height of the building (1290 ft) and solve for t:
s(t_hit) = (1/2)at_hit² = 1290
Solving for t_hit:
(1/2)(-g)t_hit² = 1290
t_hit² = -2580/g
t_hit = √(-2580/g)
(c) To determine how fast the brick is traveling when it hits the sidewalk, we need to find the velocity at t_hit.
Using the velocity function:
v(t_hit) = at_hit
Substituting the value of t_hit from part (b):
v(t_hit) = a√(-2580/g)
Note: The negative sign signifies the direction of the brick's velocity.
Therefore, the answer to part (a) and part (b) are given by:
s(t) = 0.5at²
v(t) = at
a(t) = -g
and the answer to part (c) is:
v(t_hit) = a√(-2580/g)