Answer:
[tex]\vec a = -19,096 \cdot i + 25.818 \cdot j[/tex]
Explanation:
According to the definition of the derivative of vectorial sum and the physical concept of acceleration, the required expression is:
[tex]\vec a = \frac{d v_{x}}{dt} \cdot i + \frac{d v_{y}}{dt} \cdot j + \frac{d v_{z}}{dt} \cdot k[/tex]
[tex]\vec a = (-1.88\cdot \frac{dx}{dt} + 3.94 \cdot \frac{dy}{dt})\cdot i + (1.26 \cdot \frac{dx}{dt} + 1.88 \cdot \frac{dy}{dt})\cdot j[/tex]
Where:
[tex]\frac{dx}{dt} = 0.523 - 1.88 \cdot x + 3.94 \cdot y\\\frac{dy}{dt} = - 2.44 + 1.26 \cdot x + 1.88 \cdot y\\[/tex]
The acceleration at given point is calculated as follows:
[tex]\frac{dx}{dt} = 16.193\\\frac{dy}{dt} = 2.88[/tex]
[tex]\vec a = -19,096 \cdot i + 25.818 \cdot j[/tex]
