Calculus derivatives

Question

If an object is dropped from a 237-foot-high building, its position (in feet above the ground) is given by d(t)= -16t^2+237,

where t is the time in seconds since it was dropped.
a. Compute d'(t). What units are associated with the derivative, and what does it measure?
b. When will the object hit the ground?
c. What is the velocity of the object when it strikes the ground (in mi/hr)?

Answer 1

a.  d'(t) = -32t = velocity at time t.  The units involved are ft/sec.

 

b.  The object hits the ground when the height of the object is zero.

 

     So, solve the equation d(t) = 0.

 

c.  Plug the value found in b. into the formula for d'(t). 

 

 

Answer 2

You are starting with a height function, called d(t) for distance above ground. the -16t^2 part is due to gravity and is a given thing so, no issue, Together they describe the distance above 237-16t^2, helps you see that 237 is a starting point and as the object falls, the -16t^2 subtracts from the 237 heading for the ground, d - 0.

 

OK Let's do some work d(t) = 237 - 16t^2 

a,) Compute and describe d'(t).  Well d' is a function which measures rate of change (often dy/dx, sometimes

      d/dt)(F(x,y,z,t). Here we have d'(t) = -32t measures the rate of change of distance with respect to (wrt) time. That is exactly what physicists call velocity. That means after 1 second d'(1) = -32(1) - 32 (ft./sec, the correct unit for a velocity), the 32 indicates the speed of the object when t = 1 and the - sign just means it is falling down.

b.) How long does it take to hit the floor? Well meeting the floor means the distance above (d(t)) = 0, so

 

d(t) = 0

 

237-16t^2 = 0

 

237 = 16t^2

 

237/16 = t^2                  so take the positive square root (only + root makes any sense)

 

t = sqrt(237/16) = 3.849 (For almost 4 seconds the dropped object was in "free-fall".

c.) Now the issue of how fast it collided with the floor (impact velocity or crash velocity). The object had a velocity governed by the derivative function d'(t) = -32t and "hit the ground" occurred when t = 3.849,

 

d'(3.849) - -32(3.849) = -123.168 ft./sec.. If you had a device that could measure the force of the impact, like a scale, you could measure the magnitude of the force absorbed by the collision. 

 

Good Luck

 

Dr. Dave D.