A skateboarder skates down a ramp for 3 seconds. If his initial velocity was 0.8 m/s and · his final velocity was 7 m/s, what was his acceleration?

The acceleration of the skateboarder is 2.067 m/s²

Explanation:

Given;

initial velocity of the  skateboarder, u = 0.8 m/s

final velocity of the  skateboarder, v = 7 m/s

time of motion, t = 3 s

The acceleration of the skateboarder is given as;

Therefore, the acceleration of the skateboarder is 2.067 m/s²

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

6 cos 34, 6 sin 34 trig