Answer: The magnitude of the horizontal component of the acceleration is [tex]1.732 m/s^2[/tex]
Explanation:
Acceleration of the truck = [tex]2.0m/s^2[/tex]
Angle of the slope on which truck is moving = [tex]30^o[/tex]
On splitting the resultant acceleration into vertical component and horizontal component
Vertical components = [tex]asin\theta=asin 30^o=2.0m/s^2\times \frac{1}{2}2.0m/s^2\times 0.5=1.0m/s^2[/tex]
Horizontal components =[tex]acos\theta=acos 30^o=2.0 m/s^2\times \frac{\sqrt{3}}{2}=2.0\times 0.866=1.732 m/s^2[/tex]
The magnitude of the horizontal component of the acceleration is [tex]1.732 m/s^2[/tex]
