Equations of motion

Displacement S = ut + \frac{1}{2}at^2

Final velocity v = u + at

Where, u - initial velocity, v -final velocity, t - time and a - acceleration

Given that

Initial velocity of the particle u = 10 \hat{j} m/s

Acceleration a = 8.0 \hat{i} + 2.0 \hat{j} m/s^2

a)

Displacement along the x -direction S_x = 16 m

Initial velocity of the particle u = 0 \hat{j} + 10 \hat{j}

Velocity components

velocity along the x direction u_x = 0

velocity along the y direction u_y = 10

Acceleration components

a_x = 8.0

a_y = 2.0

Using displacement to calculate the time of the particles

S_x = u_x t + \frac{1}{2}a_x t^2

16 = 0 + \frac{1}{2} (8)(t)^2

t^2 = \frac{16*2}{8}

t = \sqrt{4} = 2

Hence, After 2 seconds, the particle has moved 16 meters in the x direction.

y-coordinate of the particle

S_y = u_y t + \frac{1}{2}a_y t^2

S_y = (10)*2 + \frac{1}{2} (2)(2)^2

S_y = 20 + 4

Hence, displacement y-coordinate of the particle S_y=24\ m

b)

After 2 seconds, the particle's speed in the x direction v_x = u_x + a_x t

v_x = 0 + 8.0 (2)

v_x = 16 m/s

After 2 seconds, the particle's speed in the y direction v_y = u_y + a_y t

v_y = 10 + 2*2

v_y = 14 m/s

Magnitude of the velocity |v| = \sqrt{v_x^2 +v_y^2} = \sqrt{16^2 +14^2}

|v| = \sqrt{256 + 196} = 21.26 m/s

Hence, At 2 seconds, speed of the particle = 21.26 m/s