Question

A supersonic airplane is flying horizontally at a speed of 2610 km/h.

1. What is the centripetal acceleration of the airplane, if it turns from North to East on a circular path with a radius of 80.5 km?
2. How much time does the turn take?
3. How much distance does the airplane cover during the turn?

Answer 1

the calculated values are approximately:

- Centripetal Acceleration
`(\(a_c\)): \(6.78 \, \text{m/s}^2\)`

- Time of Turn
`(\(t\)): \(698.77 \, \text{s}\)`

- Distance Covered
`(\(d\)): \(1985.13 \, \text{km}\)`

Let's go through each part of the problem:

a. Centripetal Acceleration:

Given that the speed of the airplane v is
`\(2720 \, \text{km/h}\)` and the radius r is
`\(84.0 \, \text{km}\)`, let's calculate the centripetal acceleration
`(\(a_c\))`.

First, convert the speed from kilometers per hour to meters per second:

`\[ v = 2720 \, \text{km/h} * \frac{1000 \, \text{m}}{1 \, \text{km}} * \frac{1 \, \text{h}}{3600 \, \text{s}} \]\[ v = (2720 * 1000)/(3600) \, \text{m/s} \]`

Now, substitute the values into the centripetal acceleration formula:

`\[ a_c = (v^2)/(r) \]\[ a_c = (\left((2720 * 1000)/(3600)\right)^2)/(84 * 10^3) \]`

Calculate
`\(a_c\)`.

`\[ v = (2720 * 1000)/(3600) \, \text{m/s} \]\[ a_c = (\left((2720 * 1000)/(3600)\right)^2)/(84 * 10^3) \]\[ a_c \approx ((755.56)^2)/(84 * 10^3) \]\[ a_c \approx (570004.94)/(84 * 10^3) \]\[ a_c \approx 6.78 \, \text{m/s}^2 \]`

b. Time of Turn:

The time t taken for one complete revolution is given by the formula:

`\[ t = (2 \pi r)/(v) \]`

Substitute the known values into the formula:

`\[ t = (2 \pi * 84 * 10^3)/((2720 * 1000)/(3600)) \]`

Calculate t.

`\[ t = (2 \pi * 84 * 10^3)/((2720 * 1000)/(3600)) \]\[ t \approx (2 * 3.1416 * 84 * 10^3)/((2720 * 1000)/(3600)) \]\[ t \approx (527885.76)/(755.56) \]\[ t \approx 698.77 \, \text{s} \]`

c. Distance Covered:

The distance d covered during the turn is given by the formula:

`\[ d = vt \]`

Substitute the known values into the formula:

`\[ d = \left((2720 * 1000)/(3600)\right) * (2 \pi * 84 * 10^3)/((2720 * 1000)/(3600)) \]`

Calculate d.

`\[ d = \left((2720 * 1000)/(3600)\right) * (2 \pi * 84 * 10^3)/((2720 * 1000)/(3600)) \]\[ d \approx (755.56 * 2 * 3.1416 * 84 * 10^3)/((2720 * 1000)/(3600)) \]\[ d \approx (1501664.66)/(755.56) \]\[ d \approx 1985.13 \, \text{km} \]`

So, the calculated values are approximately:

- Centripetal Acceleration
`(\(a_c\)): \(6.78 \, \text{m/s}^2\)`

- Time of Turn
`(\(t\)): \(698.77 \, \text{s}\)`

- Distance Covered
`(\(d\)): \(1985.13 \, \text{km}\)`

Answer 2

Centripetal acceleration of this aircraft: approximately
`6.52\; {\rm m \cdot s^(-2)}`.

Distance covered during the turn: approximately
`63.2\; {\rm km}`.

Time required for the turn: approximately
`0.0242\; \text{hours}` (approximately
`87.2\; {\rm s}`.)

Step-by-step explanation:

Convert velocity and radius to standard units:

`\begin{aligned}v &= 2610\; {\rm km \cdot h^(-1)} * \frac{1\; {\rm h}}{3600\; {\rm s}} * \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &= 725\; {\rm m\cdot s^(-1)} \end{aligned}`.

`\begin{aligned} r = 80.5\; {\rm km} * \frac{1000\; {\rm m}}{1\; {\rm km}} = 8.05 * 10^(4)\; {\rm m}\end{aligned}`.

Hence, the centripetal acceleration of this aircraft:

`\begin{aligned} a &= (v^(2))/(r) \\ &= \frac{(725\; {\rm m\cdot s^(-1)})^(2) }{8.05 * 10^(4)\; {\rm m}} \\ &\approx 6.53\; {\rm m \cdot s^(-2)}\end{aligned}`.

The trajectory of the turn is an arc with a radius of
`r = 80.5\; {\rm km}` and a central angle of
`\theta = 90^(\circ) = (\pi / 4)`. The length of this arc would be:

`\begin{aligned} s &= r\, \theta \\ &= 80.5\; {\rm km} * (\pi / 4) \\ &\approx 63.2\; {\rm km}\end{aligned}`.

The time required to travel
`63.2\; {\rm km}` at a speed of
`2610\; {\rm km \cdot h^(-1)}` would be:

`\begin{aligned}t &= (s)/(v) \\ &\approx \frac{63.2\; {\rm km}}{2610\; {\rm km \cdot h^(-1)}} \\ &\approx 0.0242\; {\rm h} \\ &\approx 0.0242 \; {\rm h} * \frac{3600\; {\rm s}}{1\; {\rm h}} \\ &\approx 87.2\; {\rm s} \end{aligned}`.