## Answers

Please, post the other question separately

The acceleration of the car is: Av 160 - 60 km = 1.667 h min a = At 60 4. Remember that the average velocity is: . V +02 Vavg .

2 From the initial moment to 15 minutes: V1 = 607 V2 = V1 + at = (60) + (1.667)(15) = 85 Average velocity: 60 + 80 km Vavg= 2 = 0 h From the initial moment to 30 minutes: km V2 = 607 km V2 = V1 + at = (60) +(1.667)(30) = 110 Average velocity: 60 + 110 km Vavg==2 =857 From the initial moment to 45 minutes: km V1 = 607 V2 = va + at = (60) + (1.667)(45) = 135 ** Average velocity: 60 + 135 _ km Vavg = 2 = 97.57 From the initial moment to 60 minutes: km V60 km V2 = 60 h km v2 = V1 + at = (60) + (1.667)(60) = 160 Average velocity: km Vavg =2 =110 h 60+ 160 = 110

The distance is giving by the area below the graph: (v1 + 02) Ax = - 2 70*(15min) 60 min Ax = Vavgat From the initial moment to 15 min: km 1h = 17.5 km the initial moment to 30 min: km 1h = 42.5 km initial moment to 45 min: Ax = 97.5 the initial moment to 60 min: 1h Ax = 11 60 min .. = 110 km km (30min) 60 min 1h 19(45min) on 60 min = 73.125 km km 1 For a motion with constant acceleration: x = xo + vt + at In this case: Xe = 0 km km Vo = 60 n = min km a = 1.661hmin Then, the general equation is: x = 1t + 0.335t2 With t in minutes and x in km

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