Question

Can someone explain why they used 56.6 and 503.4 inside the square root, and not 560 and 80, which were the given speeds.

Determine the horizontal and vertical components of Vpw and Vwg Upw, x = 0 Vpw.y = -560 heulh ® Vwq, x = 80 bulle. 603 45º = 56,57 bumple Vwg, y = 80 kully. See 45° = 56.57 wulh ② Deductions : Wrong number -2 g sign No writ i one time deduction ] (Dedustións cannat lower eadı score below.o) corona

Answer 1

Solution: - = 80kmh 560 km/h first we need to find the Component of both velocities in. both direction. (x and y). Vgw, r = 80 208 45º = 56.57 km) Vgw, y = 80 Synes' = 56.57 km) Vpw, x = 0 Vplo, y = -560 kemih Vector Resultant velocity velocity will be the sai vector sum of both. velouties let the resultant velocuty be VR VR x = Vgo, x + Vpw, x = 56-57 +0. - 56.57 km/h Very = Vg1o, 2y + Vpw, y = 56-57 -5 - = -503.4 km/h - TRIF V (JR, x)2 + (WR, y)2 . (56-57) & (503.97² It happen because we need to take the Square of both components of Result vectors not the individual vectors.