Page 77 - Schaum's Outline of Differential Equations

60 APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAR 7

Substituting these values into (1), we have for the temperature of the body at any time t

Since we require T at the initial time t = 0, it follows from (4) that

7.11. A body of mass 5 slugs is dropped from a height of 100 ft with zero velocity. Assuming no air resist-
ance, find (a) an expression for the velocity of the body at any time t, (b) an expression for the position
of the body at any time t, and (c) the time required to reach the ground.

Fig. 7-5

(a) Choose the coordinate system as in Fig. 7-5. Then, since there is no air resistance, (7.5) applies: dvldt = g. This
differential equation is linear or, in differential form, separable; its solution is v = gt+c. When t=0, v = 0
2
(initially the body has zero velocity); hence 0 = g(0) + c, or c = 0. Thus, v = gt or, assuming g = 32 ft/sec ,

(b) Recall that velocity is the time rate of change of displacement, designated here by x. Hence, v = dxldt, and
(1) becomes dxldt = 32t. This differential equation is also both linear and separable; its solution is

2
But at t = 0, x = 0 (see Fig. 7-5). Thus, 0 = (16)(0) + c 1; or Cj = 0. Substituting this value into (2), we have

(c) We require t when x = 100. From (3) t = V(100) 1(16) = 2.5sec.

7.12. A steel ball weighing 2 Ib is dropped from a height of 3000 ft with no velocity. As it falls, the ball
encounters air resistance numerically equal to v/8 (in pounds), where v denotes the velocity of the ball
(in feet per second). Find (a) the limiting velocity for the ball and (b) the time required for the ball to
hit the ground.
Locate the coordinate system as in Fig. 7-5 with the ground now situated at x = 3000. Here w = 2 Ib and
32
k= 1/8. Assuming gravity g is 32 ft/sec , we have from the formula w = mg that 2 = m(32) or that the mass of the
ball is m= 1/16 slug. Equation (7.4) becomes