Answer:
a. [tex]\vec{v} = \vec{v_x} + \vec{v_y} = (20 + 4t)\hat{x} - 15 \hat{y}[/tex]
b. 25m/s
Explanation:
Let t be the time.
The velocity in the x direction that is subjected to acceleration of 4 m/s2
[tex]v_x = 20 + 4t [/tex] m/s
The velocity in the y direction
[tex]v_y = -15[/tex] m/s
The total velocity at any time, which is the combination of both x and y vectors of velocity
[tex]\vec{v} = \vec{v_x} + \vec{v_y} = (20 + 4t)\hat{x} - 15 \hat{y}[/tex]
b) at t = 5
[tex]v^2 = v_x^2 + v_y^2 = (20 + 4t)^2 + 15^2 = (20 + 4*5)^2 + 15^2 = 400^2 + 225 = 625[/tex]
[tex]v = \sqrt{625} = 25 m/s[/tex]