Answer:
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;
[tex]a = \frac{v-u}{t}\\\\a = \frac{7-0.8}{3}\\\\a = 2.067 \ m/s^2[/tex]
Therefore, the acceleration of the skateboarder is 2.067 m/s²