Question

A computer model displays the motion of a particle on a coordinate system in real time. At time t = 0, the particle is at the origin of the coordinate system and has velocity components-= 0 and yy= 6,4 m/s. The partide has acceleration components of a, = -4.0 m/s? and ay= O

Answer 1

Since the acceleration on the y-axis is 0, then the movement in the vertical direction can be modeled as a constant speed motion:

`\begin{gathered} v_y(t)=v_(0y) \\ y(t)=y_0+v_(0y)t \end{gathered}`

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:

`\begin{gathered} v_x(t)=v_(0x)+a_xt \\ x(t)=x_0+v_(0x)+(1)/(2)a_xt^2 \end{gathered}`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):

`\begin{gathered} x(t)=(1)/(2)(-4.0(m)/(s^2))t^2 \\ \\ \therefore x(t)=(-2.0(m)/(s^2))t^2 \end{gathered}`
`y(t)=(6.4(m)/(s))t`
`\begin{gathered} v_x(t)=(-4.0(m)/(s^2))t \\ \\ v_y(t)=6.4(m)/(s) \end{gathered}`

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:

`\begin{gathered} x(4.5s)=(-2.0(m)/(s^2))(4.5s)^2=-40.5m \\ \\ y(4.5s)=(6.4(m)/(s))(4.5s)=28.8m \end{gathered}`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:

`\begin{gathered} v_x(4.5s)=(-4.0(m)/(s^2))(4.5s)=-18(m)/(s^2) \\ \\ v_y(4.5s)=6.4(m)/(s) \end{gathered}`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.