Calculus Scoring Guidelines

CALCULUS AB SCORING GUIDELINES Question 3 A car is traveling on a straight road with velocity 55 ft/sec at time t = 0 For 0 <= t <= 18 seconds, the car's acceleration a(t) in ft / s * o * c ^ 2 is the piecewise linear function defined by the graph above. 0 15 (14.15) 0 (second) (a) Is the velocity of the car increasing at t = 2 seconds? Why or why not? -15 (10-15) (14-15) (b) At what time in the interval 0 <= t <= 18 other than t = 0 is the velocity of the car 55 ft /sec? Why? (c) On the time interval 0 <= t <= 18 what is the car's absolute maximum velocity, in ft/sec, and at what time does it occur? Justify your answer. (d) At what times in the interval 0 <= t <= 18 if any, is the car's velocity equal to zero? Justify your answer. (a) Since v' * (2) = a(2) and \mathfrak{a}(2) = 15 > 0 the velocity in increasing at t = 2 1: answer and reason (b) At timet 12 because v(12) - v(0) = integrate a(t) dt from 0 to 12 = 0 1:12 1: resson 2 (c) The absolute maximum velocity is 115 ft/sec at t = 6 The absolute maximum must occur at 6 or at an endpoint. v(6) = 55 + integrate a(t) dt from 0 to 6 = 55 + 2(15) + 1/2 * (4)(15) = 115 > v(0) integrate a(t) dt from 0 to 18 < 0sov(18) < v(6) 1 / t = 6 1: absolute maximum velocity 1: identifies t = 6 and 18 as candidates indicates that v increases, decreases, then increases 1: eliminates t = 18 (d) The car's velocity is never equal to 0. The absolute minimum occurs at 16 where v(16) = 115 + integrate a(t) dt from 0 to 16 = 115 - 105 = 10 > 0 1: answer 2:1 1: reason #3 @v'(t) = a(t) v' * (a) = a(a) = 15 > 0 vel. increasing at t = 2 . OR ala) is Dos. means concave upu v(2) would be increasing (0)=55 ft/sec V(12) = 55 ft/sec meed area sh * 0 = 0 (so positive & negative read to canal each other ant) integrate a(t) dt from 0 to 6 integrate a(t) dt from 6 to 12 a(15 ^ 4) + 1/2 * (4)(15) 1 2 ( 4(- 15) + (2)(- 15) 30+30 -30-30 60 30030 -60 60-60-0 integrate a(t) dt from 0 to 12 = v(12) - v(0) 0 = V(12) - 55 V(12) = 55ft / s * es ) absolute max. test pts integrate 6a(t) dt v(6) - V(0) = integrate a(t) dt from 2 to 6 V(6)-55=60 V(6)=115 ft/sec V(18) - V(0) = integrate a(t) dt from 0 to 18 t = 0, be , Me,18 slin t = 0 55ft/sec t = 6 115 ft/sec 18 25 ft/sec V(18) - 55 = integrate a(t) dt from 0 to 12 + integrate a(t) dt from 12 to 18 OS 6 V -15) (4 + 2) + 1/2 * (2)(15) Changes inc max V(18) = 55 + (- 45) + 15 abs, max occurs et t = 6 a5ft/sec occurs at rel, max t-to er an endpt t=0 X = 1Y abs, mat occurs at t = 6 ) + 115fHsec V(l6)-V(0)= int 0 ^ ln a(t)dt @abs min occurs t = 16 j * (lb)' - V(0) = integrate a(t) dt from 0 to 6 + integrate a(t) dt from 6 to 16 To's o vel A