Question

please be very clear and detailed in showing each step taken to solve. please don't skip any steps either. thank you!

1. (25pt) The velocity for a two-dimensional flow is defined by V = ((0.5x3yt)i - (2x - yt)j)m/s, where x and y are in meters. Determine the magnitudes of the velocity and acceleration of a particle located at the point x = 1m, y = 2m.

Answer 1

Given - V = (0.5x3y+) i (2x - Y+) j m where, t = time, magnitude of velocity, at xrim, y= 2m, V = (0.5) (imje(2m) Its i -261) – 24}j OR ✓ ti_21-tjj Iol= magnitude of velocity, 171 = Ve +4211-+)* or 157 = + + 461++-2+) Or lūl = 5+2-84+4 mis Here relocity at time =t see. NOW, Acceleration (9) = do at acc. in x di reetion, ax u.au 2x + vou 09 + ou at ay = Scanned with CamScanner Ovi ax + var' ay Y av' at

2 Here, ū - vi tv'j 0.5x3y+ va - (2x - Y+) ro, ax = U Bu ya + vou ay + OU ət au 1.5*?y+ ax au 11 0.5x3+ au 2+ 0.5x3y so, At X=im, y = 20, Qe = (0.5 x3y +) (1.5x2y+) - (2x-Y+) (0.5*34) + 0.5x3y ori 3+2 2 (t-1) (0.5+) + 1 OY ex = 3+2 - It-t?) + 1 (4t?-++) i ori ax = Seanned with CamScanner

Similarly, 3 ay = u ovi + v'av! ax + ay av! Ət əv' Әҳ ul -2 əv' əy t av' O+ = y ΟΥ, ay = (0.5x3y+) 62) - (2x-Y+) + +y butting X=1, y = 2 ay = - 2t 2 (1-t).t + 2 Or ay - 2+2-47 +2 or ay = 21 +- 24 +) j = 2 (t-1) 2j So, ão xi tayi ă 2 = (4+-++i)i + 2(t-1); lal - ✓lut? +++)² + 4H-114 Here, velocity & acceleration ý given of time t = t see. CSScanned with CamScanner