Question

1. An airplane travelling at 350 m/s [55 degrees W of N], takes 54 seconds to change its velocity to 150 m/s [30 degrees S of E]. What is the average acceleration over this time intervalThe answer I got a=-4.13413 m/s^2 [W 71.496 N]

Answer 1

Answer:

The average velocity over the given interval = 4.13 m/s

Explanation:

The initial velocity, v₀ = 350 m/s [55 degrees W of N]

The final velocity, vf = 150 m/s [30 degrees S of E]

Resolve the initial velcity to the x direction

$\begin{gathered} v_{0x}=350\cos 55 \\ v_{0x}=200.75\text{ m/s} \end{gathered}$

Resolve the initial velocity to the y direction

$\begin{gathered} v_{oy}=350\sin 55 \\ v_{oy}=286.7\text{ m/s} \end{gathered}$

Resolve the final velocity to the x-direction

$\begin{gathered} v_{fx}=150\cos 30 \\ v_{fx}=129.9\text{ m/s} \end{gathered}$

Resolve the final velocity to the y-direction

$\begin{gathered} v_{fy}=150\sin 30 \\ v_{fy}=75\text{ m/s} \end{gathered}$

The acceleration in the x-direction

$\begin{gathered} a_x=\frac{v_{fx}-v_{0x}}{t} \\ a_x=\frac{129.9-200.75}{54} \\ a_x=\frac{-70.85}{54} \\ a_x=-1.31m/s^2 \end{gathered}$

The acceleration in the y-direction

$\begin{gathered} a_y=\frac{v_{fy}-v_{0y}}{t} \\ a_y=\frac{75-286.7}{54} \\ a_y=-3.92\text{ m/s} \end{gathered}$

Find the resultant acceleration

$\begin{gathered} a=\sqrt[]{a^2_x+a^2_y} \\ a=\sqrt[]{(-1.31)^2+(-3.92)^2} \\ a=\sqrt[]{17.0825} \\ a=4.13\text{ m/s} \end{gathered}$

The average velocity over the given interval = 4.13 m/s