Question

Assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.

Answer 1

We are given that a jet is traveling with a speed of 78.6 m/s and travels a distance of 919m. We are asked to determine the constant acceleration when the jet stops. To do that we will use the following formula:

`v^2_f=v^2_0+2ax`

Where:

`\begin{gathered} v_f=\text{ final speed} \\ v_0=\text{ initial speed} \\ a=\text{ acceleration} \\ x=\text{ distance traveled} \end{gathered}`

Since the jet stops, this means that the final speed is zero. We will solve for the acceleration "a" in the formula. First, we will eliminate the term for the final speed since it is zero:

`0=v^2_0+2ax`

Now we will subtract the initial speed squared from both sides:

`-v^2_0=2ax`

Now we will divide by "2x" from both sides:

`(-v^2_0)/(2x)=a`

Now we replace the known values:

`(-(78.6(m)/(s))^2)/(2(919m))=a`

Solving the operations:

`-3.36(m)/(s^2)=a`

Therefore, the magnitude of the acceleration is 3.36. Since the jet is deaccelerating in the direction due south, the direction of the acceleration is due north.