Sin ( 1291) 756 - 433 m (N] the w component is 250 m [ w] the Newponent is 43 3 m [ N] 6) A student runs at 4.5 m/s [250 S of W] for 3.0 minutes. Then he jogs at 3.5 m/s [350 S of E] for 4.1 minutes. Given! VectorI and vector ? Required: Average speed ( a) What was his average speed? +1 = 2465 45+=1 1265 V = 2 V = 3.92 mls AT, = 4. 1 minx GOS = 246 +2 = 1805 AT2 = 3. 0 min x 605 = 180 d = 810 m+ 861 m V = 1671 m ad= ( 4. 5 m 15) ( 1805)-210 m Ad ( 3. 5 m / 5) (2465) = 861m d= 1671 m 4265 the avg speed is 3.92 mls b) What was his final displacement? Given it. = 2465 V. = 4. Smals [zsosol wi ] + 2 = 1805 V2 = 3. 5mils [ 350 5 of E] V = 14 . 5 m / 5 ) ( cos 25# (W]) V = ( 3. 5 m/s) ( cos ( 35 ) (E]) Required: final displacement V = 4. 1 m/s v = 2.9mls Let Nand E be positive V = (4. 5 mils) (sin 25) v= (3. 5 m/s) (sin) 350) [5] and sand w be negative V=19m /s [s] V = - 2,0mis [s] V+ = -14, 1 wls+ ( claimls) vr=- 2:9mls + (-2. omls) = 232 mls [ W] = 01.9 mls [s] V = 1 / + 2. 2 mls ) 2 + ( - 1 . 9 mls ) 2 = 4.1 m/s