Since the acceleration on the y-axis is 0, then the movement in the vertical direction can be modeled as a constant speed motion:
[tex]\begin{gathered} v_y(t)=v_{0y} \\ y(t)=y_0+v_{0y}t \end{gathered}[/tex]On the other hand, since the acceleration on the x-axis is different from 0 and it is a constant acceleration, then, the horizontal movement of the particle can be modeled as a uniformly accelerated motion:
[tex]\begin{gathered} v_x(t)=v_{0x}+a_xt \\ x(t)=x_0+v_{0x}+\frac{1}{2}a_xt^2 \end{gathered}[/tex]Part a)Since the particle is at the origin at t=0, then x₀=0 and y₀=0.
Replace v_0x=0, v_0y=6.4m/s and a_x=-4m/s^2 into the equations to find the expressions for x(t), y(t), v_x(t) and v_y(t):
[tex]\begin{gathered} x(t)=\frac{1}{2}(-4.0\frac{m}{s^2})t^2 \\ \\ \therefore x(t)=(-2.0\frac{m}{s^2})t^2 \end{gathered}[/tex][tex]y(t)=(6.4\frac{m}{s})t[/tex][tex]\begin{gathered} v_x(t)=(-4.0\frac{m}{s^2})t \\ \\ v_y(t)=6.4\frac{m}{s} \end{gathered}[/tex]Evaluate the expressions for x(t) and y(t) at t=4.5s to find the x and y positions of the particle at t=4.5s:
[tex]\begin{gathered} x(4.5s)=(-2.0\frac{m}{s^2})(4.5s)^2=-40.5m \\ \\ y(4.5s)=(6.4\frac{m}{s})(4.5s)=28.8m \end{gathered}[/tex]Part b)Replace t=4.5s into the expressions for the velocities of the particle to find v_x and v_y at t=4.5s:
[tex]\begin{gathered} v_x(4.5s)=(-4.0\frac{m}{s^2})(4.5s)=-18\frac{m}{s^2} \\ \\ v_y(4.5s)=6.4\frac{m}{s} \end{gathered}[/tex]Part c)Initially, the horizontal component of the velocity was 0 while the vertical component of the velocity was 6.4m/s. After 4.5 seconds, the vertical component of the velocity stays the same but the horizontal component changes to -18m/s. The magnitude of the velocity increases because the magnitude of one of the components increases while the other remains the same.
Therefore, the answers are:
Part a)
x = -40.5 m
y = 28.8 m
Part b)
v_x = -18 m/s
v_y = 6.4 m/s
Part c)
The particle's speed increases with time.
