Question

M10. A river is 1.0 km wide, and flows at 10 km/h. Points A and B are on opposite banks, directly across from each other, while point C is 1.0 km upstream from A. A motorboat which can move at 25 km/h in still water travels from A to B, then from B to C. What is the minimum possible time for this trip? (7.64 min)

Answer 1

Ans Page No. Date : mo itm 10km/h 1 km AC = ' km AB -1 km by Pythagoras 8C = 60% CABI - Be = (c. ²7 (172 = 12 km boat's speed =25m/h and siver speed =10km/h A to B (a) when boat goes from Point then speed distance ime. a time= distance speed t = 1km I 25km t = 0.04 hours

Page No Date: when you goes from B to c tesinys B 12 m 70-45 km Ytoys C 1km " tano = AC=1=1 AB I tano=1 oztañ(1) =450=45° Now, Lets divide time in x and y component for y-axis tosys distance 520km og speed 25 sinys to 005656 seconds for of axis component of time - the 0.05656 seconds for xo-axis - ta - distance = 12 km - 5 km speed (25-10 TS here speed = 15km/h because sirer is also following in ex-axis direction Copposite direction speed decreases = 25 40 = 15 km/h

Page No. Date : = 12 - 1.414 15 15 tax sings - 1.414 - 1 15xt is tar sin45= 1 =0.06666 So Now taking magnitude of x any Component 5 et (tag.cosysji (toasinys) poc CO.06666927005656)? = 0.08737 sec I and t, (tAB) = 0.04 see total tapt toe = 0,08737 +0.04, TABt toc = 0.12737 hour - I hour= 60 min so = 0.12737 X 60 min - = 7.64 min Total time taken by bought is 7.64 min