Problem 44 - College Algebra and Calculus: An Applied Approach

Question

A brick becomes dislodged from the top of the Empire State Building (at a height of 1250 feet) and falls to the sidewalk below. (a) Write the position, velocity, and acceleration functions of the brick. (b) How long does it take the brick to hit the sidewalk? (c) How fast is the brick traveling when it hits the sidewalk?

Step-by-Step Solution

Answer

The position, velocity, and acceleration functions for the brick are \( s(t) = 1250 - 16t^2 \), \( v(t) = -32t \), and \( a = -32 \) ft/sec² respectively. The time it takes for the brick to hit the sidewalk is 25 seconds. The brick is traveling at 800 ft/sec when it hits the sidewalk.

Step 1: Position, velocity, and acceleration functions

Firstly, the position, velocity, and acceleration functions need to be written for the falling brick. In this case, the acceleration will be -32 feet/sec^2 (the acceleration due to gravity), and the initial velocity (V0) will be 0 as the brick was not thrown but simply dislodged.The general form of the position and velocity functions are: \( s(t) = s_0 + v_0 t + \frac{1}{2} a t^2 \) (position) \( v(t) = v_0 + at \) (velocity)Given \( s_0\) = 1250 feet (initial height), \( v_0\) = 0 feet/sec (initial velocity), and \( a\) = -32 feet/sec² (acceleration due to gravity), insert the values into the equations to get:\( s(t) = 1250 - 16t^2 \)\( v(t) = -32t \)And acceleration is a constant \( a = -32 \) feet/sec²

Step 2: Time to hit the sidewalk

The time it will take for the brick to hit the sidewalk can be calculated by setting the position function equal to zero and solving for t as follows:\( 0 = 1250 - 16t^2 \) Solving this equation gives you \( t = \sqrt{\frac{1250}{16}} = 25\ sec \).

Step 3: Speed of the brick when hitting the sidewalk

To find out how fast the brick will be travelling when it hits the sidewalk, substitute the time t=25 sec in the velocity function v(t), \( v(25) = -32 * 25 = -800 \) ft/sec.The speed is the absolute value of the velocity, thus \( \ |v(25)| = 800 \) ft/sec without the negative sign. The negative sign merely indicates downward motion.