Schaum's Outline of Theory and Problems of Beginning Calculus

140 RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY [CHAP. 18When t = 2, s = 256,The initial velocity was 160 feet per second, so that256 = 00(2) - 16(2)2256 = 200 - 64320 = 20,160 = 00s = 160t - 16t2 and U = 160 - 32tTo find the time when the maximum height is reached, set D = 0,0 = 160 - 32t32t = 160t = 5 secondsTo find the maximum height, substitute t = 5 in the formula for s,s = 160(5) - 16(5)2 = 800 - 16(25) = 800 - 400 = 400 feet18.3 A car is moving along a straight road according to the equation=f(t) = 2t3 - 3t2 - 12tDescribe its motion by indicating when and where the car is moving to the right, and when andwhere it is moving to the left. When is the car at rest?We have U =f'(t) = 6t2 - 6t - 12 = 6(t2 - t - 2) = 6(t - 2)(t + 1). The key points are t = 2 andt = - 1 (see Fig. 18-5).-I 2 tFig. 18-5(i) When t > 2, both t - 2 and t + 1 are positive. So, U > 0 and the car is moving to the right.(ii) As t moves from (2, CO) through t = 2 into (-1, 2), the sign of t - 2 changes, but the sign of t + 1remains the same. Hence, U changes from positive to negative. Thus, for - 1 < t < 2, the car is movingto the left.(iii) As t moves through t = - 1 from (- 1, 2) into (- a, - l), the sign of t + 1 changes but the sign oft - 2 remains the same. Hence, U changes from negative to positive. So, the car is moving to the rightwhen t < - 1.When t = -1,When t = 2s = 2(- 1)3 - 3(- 1)2 - 12(- 1) = -2 - 3 + 12 = 7s = 2(2)3 - 3(2)2 - 12(2) = 16 - 12 - 24 = -20Thus, the car moves to the right until, at t = - 1, it reaches s = 7, where it reverses direction and moves leftuntil, at t = 2, it reaches s = -20, where it reverses direction again and keeps moving to the right thereafter(see Fig. 18-6).Fig. 18-6