Question

2. A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration ā=(4.0 m/s-) +(8.0 m/s2);. At time 1 = 0, the velocity is (4.00 m/s)ì. What are the (a) magnitude and (b) angle of its velocity when it has been displaced by 16.0 m parallel to the x axis?​

Answer 1

Answer:

The acceleration in the x and y directions is ax = 5m/s2 and ay = 7m/s2 respectively velocity in x and y direction ux = 4m/s and uy = 0 x = 12m the final velocity in x dir. v2 -u2 = 2as v2 - 16 = 2*5*12 or v2 = 120+16 = 136 or v = sqrt(136) = 11.66m/s Now time taken to gain this velocity t = (v - u) / a = (11.66 - 4) / 5 = 1.53sec The velocity in y dir. vy = u + at = 0 + 7*1.53 = 10.73 m/s...