please help with this question

Question

A miniature quadcopter is located at x_i = 2.50 m

 and y_i = −5.70 m

 at t = 0

 and moves with an average velocity having components v_av, x = 2.20 m/s

 and v_av, y = −2.50 m/s.

 What are the x-coordinate and y-coordinate (in m) of the quadcopter's position at t = 1.10 s?

(a)

x-coordinate

 m

(b)

y-coordinate

 m

Answer 1

As the component of velocity in the x direction is 2.20 m/s, and it travels for 1.10 s, it moves 2.42 meters in that period of time. As its initial position on the x axis is 2.50 m, the final position on the x axis is 2.50 + 2.42 = 4.92 m on the x axis.

The velocity in the negative y direction is -2.5 m/s for 1.10 s; therefore, the final distance travelled in the negative y direction is 1.10 x -2.50 = - 2.75 m. The final position on the y axis is -2.75 +(-5.70) = - 8.45 m.

The final coordinates are x = 4.92 m and y = -8.45 m.