Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 235: 1

1a) v ( t ) = 3t² - 16t + 24 1b) v ( 1 ) = 11 feet per second 1c) Particle is never at rest. 1d) Since the particle moves forward to start and is never “at rest”, the particle is always moving in a positive direction. 1e) The particle travels 72 feet in 6 seconds. 1f) photo of 1f) sketch included 1g) a ( 1 ) = - 10 feet per second per second 1h) photo of 1h) sketch included 1i) The particle is speeding up after 8/3 seconds, ≈ 2.7 seconds. 1i) The particle is slowing down when 0 ≤ t < 8/3 , from zero seconds to approximately 2.7 seconds.

1) s = f ( t ) = t³ - 8t² + 24t a) Find velocity at time ( t ). s ‘= f ‘( t ) = v ( t ) = 3t² - 16t + 24 1a) v ( t ) = 3t² - 16t + 24 b) What is the velocity after 1 second? v ( 1 ) = 3( 1 )² - 16( 1 ) + 24 v ( 1 ) = 3 - 16 + 24 1b) v ( 1 ) = 11 feet per second c) When is the particle at rest? When is velocity zero? v ( t ) = 3t² - 16t + 24 Looks suspicious, check the discriminant. 3t² - 16t + 24 = ax² + bx + c a = 3 b = ( - 16 ) c = 24 x= \frac{-b±√b^{2}-4ac}{2a} ← quadratic equation -b± \sqrt b^{2}-4ac ← discriminant \sqrt ( -16² - 4( 3 )( 24 )) \sqrt (256-288) \sqrt (-32) = no real x-value can equal zero 1c) Particle is never at rest. d) When is the particle moving in the positive direction? When is velocity greater than zero? 1d) Since the particle moves forward to start and is never “at rest”, the particle is always moving in a positive direction. e) Find the total distance traveled during the first 6 seconds. s = f ( t ) = t³ - 8t² + 24t s = f ( 0 ) = ( 0 )³ - 8( 0 )² + 24( 0 ) = 0 s = f ( 0 ) = 0 s = f ( 6 ) = ( 6 )³ - 8( 6 )² + 24( 6 ) s = f ( 6 ) = 216 – 288 + 144 s = f ( 6 ) = 72 feet f ( 6 ) - f ( 0 ) = 72 feet 1e) The particle travels 72 feet in 6 seconds. f) Draw a diagram like Figure 2 to illustrate the motion of the particle. 1f) photo of 1f) sketch included g) Find the acceleration at time ( t ) and after 1 second. v ( t ) = 3t² - 16t + 24 v ‘( t ) = a ( t ) = 6t - 16 a ( t ) = 6t - 16 a ( 1 ) = 6( 1 ) – 16 1g) a ( 1 ) = - 10 feet per second per second 1h) photo of 1h) sketch included i) When is the particle speeding up? When is it slowing down? The particle is speeding up when v ( t ) and a ( t ) have the same signs. 1c) Particle is never at rest. 1d) Since the particle moves forward to start and is never “at rest”, the particle is always moving in a positive direction. v ( t ) always positive a ( t ) = 6t – 16 6t – 16 > 0 3t – 8 > 0 3t > 8 t > 8/3 Velocity is positive before 8/3 seconds. Acceleration is negative before 8/3 seconds. v ( t ) and a ( t ) have different signs Velocity is positive after 8/3 seconds. Acceleration is positive after 8/3 seconds. v ( t ) and a ( t ) have same signs 1i) The particle is speeding up after 8/3 seconds, ≈ 2.7 seconds. 1i) The particle is slowing down when 0 ≤ t < 8/3 , from zero seconds to approximately 2.7 seconds.