a particle starts from the origin at t0 with a velocity of 2

a particle starts from the origin at t=0 with a velocity of (2.0i+4.0j)m/s and moves in the xy plane with a constant acceleration of 5.83m/s^2 directed 59 degree below the x axis. what is the speed of the particle at t=2.0s

Solution

a = 5.83 m/s^2 59 deg below x axis.

a = 5.83 ( cos59i - sin59j) = 3i - 5j m/s^2

initial velocity, u = 2i + 4j
applying

v = u + at

v = (2i + 4j) + (3i - 5j)(2) = (2 + 6)i + (4 - 10)j

v = 8i - 6j m/s

Speed = sqrt(8^2 + 6^2) = 10 m/s ......Ans