Answer:
[tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]
Magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex]
Explanation:
[tex]t_1=2\ \text{s}[/tex]
[tex]v_x=1\ \text{m/s}[/tex]
[tex]v_y=3\ \text{m/s}[/tex]
[tex]t_2=2.5\ \text{s}[/tex]
[tex]v_x=4\ \text{m/s}[/tex]
[tex]v_y=3\ \text{m/s}[/tex]
Average acceleration in the different axes
[tex]a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2[/tex]
[tex]a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2[/tex]
The components of the acceleration is [tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]
The magnitude of acceleration
[tex]a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2[/tex]
Direction
[tex]\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}[/tex]
The magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex].
