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