Question

18) A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration . At time t = 0, the velocity is . What are the (a)(significant digits of 3) magnitude and (b)(significant digits of 3) angle of its velocity when it has been displaced by 12.0 m parallel to the x axis?

Please include units

Answer 1

Acceleration of pebble is a = (5.02 + 7.09) m/s2 At time t = 0, velocity of pebble is Ū; = 5.02 m/s ā=azî + ay] =(5.0 +7.09)= a; = 5.0, ay = 7.0 ū; = Viri + Viyj = 5.0 +0j → Vir = 5.0, viy = 0 Initially the position of pebble is Fi = 01 +0j w ;= 0, y = 0 When the pebble is displaced through a distance of 12 m parallel to x axis, final position of pebble is f = 01 + 12 => If = 12, y = y If = t; + Virt + -at 12.0 = 0 + 5.0t +-5.0 * t2 =t= 1.0483 s yg = i + vist + žart? = yg = 0 +0 ++7.0 + 1.4083° = 6.9416m Ufr = Vix + art = 5.0+ 5.0 * 1.4083 = 12.0415 m/s Ufy = Viy + ayt = 0 + 7.0 * 1.4083 = 9.8581 m/s Magnitude of final velocity Up = Votz + vžy = V12.04152 + 9.85812 = 15.6 m/s b) = tan-11 stan -119.8581 Angle 12.0415 ) = 39.3° counter clockwise with positive X axis.