the height of a falling object is given by h(t) = 400 − 39t − 157e−t/4

Question

the height of a falling object is given by

h(t) = 400 − 39t − 157e−t/4

with h in meters and t in seconds. Compute the velocity at the instant when t = 2.0 seconds. Your answer must be accurate to one decimal place.

dh

dt t=2.0 = ___________

Answer 1

We know that h(t) is representing the height as a function. Another way to call height, in a way you might have heard in class, is position. To turn h(t) into v(t), also known as position function into velocity function, we must take the derivative in respect to time (t). I’m also assuming the stated question is:

h(t)=400-39t-157e^(-t/4)

Therefore:

h(t)=400-39t-157e^(-t/4)

dh=d/dt(400-39t-157e^(-t/4))

dt=d/dt(400)-d/dt(39t)-d/dt(157e^(-t/4))

Taking each derivative separately will allow easier simplification:

d/dt(400)=0

Note: derivative of a constant is 0

d/dt(39t)=39

Note: used the power rule, which eliminates t

d/dt(157e^(-t/4))=-(157/4)e^(-t/4)

Note: used the chain rule. This results in taking the derivative of (-t/4) and multiplying it on the constant 157.

Now we can substitute the results and simplify, note that the + or - remain true to original position function until simplification is done:

=> dt=0-39-[-(157/4)e^(-t/4)]

=> dt=-39+(157/4)e^(-t/4)

dt=the velocity function known as v(t)

Solving this velocity function for the velocity when t=2.0 s is done by substituting 2.0 in for t:

v(t)=-39+(157/4)e^(-2.0/4)

=> dt=-15.2 m/s