121616949-math.198 - πt is 5 i.e when πt = 7 π 6 11 π 6...

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184 Chapter 9 Applications of Integration while the total distance traveled in the first 4 seconds is Z 2 0 ( - 9 . 8 t + 19 . 6) dt + fl fl fl fl Z 4 2 ( - 9 . 8 t + 19 . 6) dt fl fl fl fl = 19 . 6 + | - 19 . 6 | = 39 . 2 meters, 19 . 6 meters up and 19 . 6 meters down. EXAMPLE 9.6 The acceleration of an object is given by a ( t ) = cos( πt ), and its velocity at time t = 0 is 1 / (2 π ). Find both the net and the total distance traveled in the first 1.5 seconds. We compute v ( t ) = v (0) + Z t 0 cos( πu ) du = 1 2 π + 1 π sin( πu ) fl fl fl fl t 0 = 1 π ( 1 2 + sin( πt ) ) . The net distance traveled is then s (3 / 2) - s (0) = Z 3 / 2 0 1 π 1 2 + sin( πt ) dt = 1 π t 2 - 1 π cos( πt ) ¶fl fl fl fl 3 / 2 0 = 3 4 π + 1 π 2 0 . 340 meters. To find the total distance traveled, we need to know when (0 . 5 + sin( πt )) is positive and when it is negative. This function is 0 when sin(
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Unformatted text preview: πt ) is-. 5, i.e., when πt = 7 π/ 6, 11 π/ 6, etc. The value πt = 7 π/ 6, i.e., t = 7 / 6, is the only value in the range 0 ≤ t ≤ 1 . 5. Since v ( t ) > 0 for t < 7 / 6 and v ( t ) < 0 for t > 7 / 6, the total distance traveled is Z 7 / 6 1 π ± 1 2 + sin( πt ) ¶ dt + fl fl fl Z 3 / 2 7 / 6 1 π ± 1 2 + sin( πt ) ¶ dt fl fl fl = 1 π ± 7 12 + 1 π cos(7 π/ 6) + 1 π ¶ + 1 π fl fl fl 3 4-7 12 + 1 π cos(7 π/ 6) fl fl fl = 1 π ˆ 7 12 + 1 π √ 3 2 + 1 π ! + 1 π fl fl fl 3 4-7 12 + 1 π √ 3 2 . fl fl fl ≈ . 409 meters....
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