Page 90 - Calculus Demystified

Foundations of Calculus
CHAPTER 2

When f (t) represents velocity, then sometimes we calculate another derivative— 77
(f ) (t)—and this quantity denotes the rate of change of velocity, or acceleration.

In specialized applications, even more derivatives are sometimes used. For
example, sometimes the derivative of the acceleration is called jerk and sometimes
the derivative of jerk is called surge.
EXAMPLE 2.20
2
The position of a body moving along a linear track is given by p(t) = 3t −
5t + 7 feet.Calculate the velocity and the acceleration at time t = 3 seconds.

SOLUTION
The velocity is given by

p (t) = 6t − 5.

At time t = 3 we therefore find that the velocity is p (3) = 18 − 5 = 13 ft/sec.
The acceleration is given by the second derivative:

p (t) = (p ) (t) = (6t − 5) = 6.
2
The acceleration at time t = 3 is therefore 6 ft/sec .
Math Note: As previously noted, velocity is measured in feet per second (or
ft/sec).Acceleration is the rate of change of velocity with respect to time; therefore
2
acceleration is measured in “feet per second per second” (or ft/sec ).
EXAMPLE 2.21

A massive ball is dropped from a tower. It is known that a falling body
descends (near the surface of the earth) with an acceleration of about 32
ft /sec. From this information one can determine that the equation for the
position of the ball at time t is
2
p(t) =−16t + v 0 t + h 0 ft.
Here v 0 isthe initial velocity and h 0 isthe initial height of the ball in feet. 1
Also t is time measured in seconds. If the ball hits the earth after 5 seconds,
then determine the height from which the ball isdropped.

SOLUTION
Observe that the velocity is

v(t) = p (t) =−32t + v 0 .
Obviously the initial velocity of a falling body is 0. Thus
0 = v(0) =−32 · 0 + v 0 .

1 We shall say more about this equation, and this technique, in Section 3.4.