1 Let s t ), v ( t ), and a ( t ) denote the functions for distance, velocity, and acceleration at time t , respec- ( t = 0 be at 2:45 a.m., so = 1 / 2 is at 3:15 a.m. We’re given that s (0) = 0 miles, v (0) = 0 tively. Let t . The goal is to (1 2) = 160 miles, and we’re given that a ( s ) = A for some constant A / mph, and t v (1 / 2). find Recall ∫ ′ ) = s C ( t ) v v ( t ) dt = s ( t ) + ( t ∫ ′ ( v ( a t ) ) = a ( t ) dt = v ( t ) + C. t Hence, ∫ ∫ ( t ) dt = A dt a At + C = v t ) . = ( v (0) = 0, C = 0, so v ( t Since At , where A needs to be determined. Note that after finding A , ) = we can then evaluate v (1 / 2) to get the answer. The problem says we need to use anti-derivatives at some point, so here it goes: Note that s ( t ) is the anti-derivative of v ( t ), so by the FTC: ∫ 1 / 2 2) t dt = s v / ) − s (0) ( (1 0 = 160 miles , ( ) = At , t v and since we also know ∫ ∫ 1 / 2 1 / 2 dt = v ( t ) At dt 0 0 ∣ / 1 2 ∣ A 2 ∣ t = ∣ 2 0 A = miles . 8 2 A/ 8 = 160, so A = 1280 miles/hr Setting these two quantities equal to one another gives . Thus, v ( ) = 1280 t , so v (1 / 2) = 640 miles/hr is our answer. t Check! We have 2 t ) = 640 t ( s v ( t ) = 1280 t a ( t ) = 1280 . ′ ′ ) = t ) = s Note that ( t ) and a ( t v v ( ( t ), and that s (0) = 0, s (1 / 2) = 160, and v (0) = 0.