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

Step-by-step 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=(v_(fx)-v_(0x))/(t) \\ a_x=(129.9-200.75)/(54) \\ a_x=(-70.85)/(54) \\ a_x=-1.31m/s^2 \end{gathered}`

The acceleration in the y-direction

`\begin{gathered} a_y=(v_(fy)-v_(0y))/(t) \\ a_y=(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