Example Problem #2.33 x E M v t a t v 0 t1t1 t2t2 t3t3 20 m/s 2 –20 m/s 2 t1t1 t2t2 t3t3 0

Example Problem #2.33 (a) max. speed attained when acceleration = 0  constant speed achieved (aqua region)  v = v 0 + at  look at times when t = 0 (v 0 = 0) and t = t 1 (v = max. and const.)  v = at 1  a = 20.0 m/s 2  t 1 = 15 min = 900 s  v = (20.0 m/s 2 )(900 s) = 18,000 m/s = 18.0 km/s (b) 1 st leg of journey (represented by red line in graphs): x – x 0 = ½(v 0 + v)t  x 0 = 0, v 0 = 0  x = x 1 = vt/2  v = 18.0 km/s  t = 900 s  x 1 = 8100 km 3 rd leg of journey (yellow line): x – x 0 = ½(v 0 + v)t  x 3 – x 2 = ½(v 2 + v 3 )t  v 2 = 18.0 km/s, v 3 = 0, t = 900 s  x 3 – x 2 = 8100 km  leg 1 + leg 3 = 16,200 km

Example Problem #2.33 (b continued) leg 2 distance = 384,000 km – 16,200 km = 367,800 km  fraction of total distance = leg 2 dist. / total dist. = 367,800 km / 384,000 km = (c) Find time ship was traveling at constant speed (leg 2, aqua region): x – x 0 = ½(v 0 + v)t  x 2 – x 1 = ½(v 1 + v 2 )t  x 1 = 8100 km, x 2 = 384,000 km – 8100 km = 375,900 km, v 1 = 18.0 km/s, v 2 = 18.0 km/s  t = 2(x 2 – x 1 ) / (v 1 + v 2 ) = 20,433 s = t leg2  t total = t leg1 + t leg2 + t leg3 = 900 s + 20,433 s s = 22,233 s = min = 6.18 hr