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Honors 2201-4 Homework 9 Solutions Fall 2003 Section 3.3 (2) We set the initial location s(0) = 0, the initial velocity v(0) = 0 and define the downward direction to be positive. We know that the acceleration due to gravity is a(t) = g = 32.17f t/s2. Therefore, dv =g v(0) = 0 dt v = gt + c1 ⇒ c1 = 0 v(t) = gt ds = gt s(0) = 0 dt 1 s = gt2 + c2 ⇒ c2 = 0 2 1 s(t) = gt2 2 We must find the time, tg , when the marble hits the ground, or s(tg ) = 4. 1 4 = gt2g 2 s 8 tg = ≈ 0.499 g The marble takes about half a second to hit the ground. Its velocity when it hits is v(tg ), v s ! 8 g =g s 8 q = 8g ≈ 16 g The marble is moving at 16f t/sec when it hits the ground. (22) dy = 3t2 dt y(2) = 5 y = t3 + c 5 = 23 + c ⇒ c = 5 − 8 = −3 y(t) = t3 − 3 1 (25) dy = 2t + 2e−2t y(0) = 3 dt y = t2 − e−2t + c 3 = 0 − e0 + c = c − 1 ⇒ c = 4 y(t) = t2 − e−2t + 4 (27) dy = 1 + 3t2 + 3e−t y(0) = 1 dt y = t + t3 − 3e−t + c 1= 0+0−3+c ⇒c =4 y(t) = t + t3 − 3e−t + 4 (35) We are given a(t) = 1.4f t/sec2 and v(0) = 0. We will set s(0) = 0 and assume the direction of travel is positive. Therefore, dv = 1.4 v(0) = 0 dt v = 1.4t + c1 ⇒ c1 = 0 v(t) = 1.4t ds = 1.4t s(0) = 0 dt 1.4 2 t + c2 ⇒ c2 = 0 s= 2 s(t) = .7t2 where time, t, is measured in seconds. At the end of one minute the car is traveling at v(60) = 1.4(60) = 84f t/sec ≈ 57mi/hr. During that time the car travels s(60) = .7(60)2 = 2520f t. (39) We are told that the initial velocity is 285k/hr. Converting this to seconds we have v(0) = (285/3600)k/sec and v(28) = 0, where time, t, is measured in seconds. Let s(0)=0 2 and assume that the direction of travel is the positive direction. We are also told that the deceleration rate is proportional to time t. Therefore, the acceleration a(t) = −kt, where k > 0 since the the plane is slowing down. Then, dv = −kt dt v(0) = 285 3600 k v = − t2 + c1 2 k 2 285 285 = (0) + c1 ⇒ c1 = 3600 2 3600 k 2 285 v(t) = − t + 2 3600 We know that the plane stopped after 28 seconds, so v(28) = 0. Therefore, 285 k 0 = − (28)2 + 2 3600 285(2) k= ≈ .0002 3600(28)2 Now, we must find the distance traveled in the first 28 seconds. First, we find the equation for the location over time using the initial condition s(0) = 0. s(28) = − 1 6 ds k 285 = v(t) = − t2 + dt 2 3600 285 k 3 t + c2 ⇒ c2 = 0 s=− t + 6 3600 285 k t s(t) = − t3 + 6 3600 ! 285(2) 285 (28)3 + (28) ≈ 1.48 2 3600(28) 3600 The plane travels approximately 1.5 kilometers before stopping. 3