Question

A boy is twirling a model airplane on a string 4 feet long. If he twirls the plane at 0.5 revolutions per minute, how far does the plane travel in 4 minutes

Answer 1

Answer:

Airplane covers a distance of approximately 50.265 feet in 4 minutes.

Step-by-step explanation:

From Rotation Physics we know that angular speed is the number of revolutions done by a particle at a given time. If the boy is twirling a model airplane at constant speed, then we calculate such variable as follows:

$\dot n = \frac{n}{t}$ (Eq. 1)

Where:

$\dot n$ - Angular speed, measured in revolutions per minute.

$n$ - Number of revolutions done by airplane, measured in revolutions.

$t$ - Time, measured in minutes.

Now we clear the number of revolutions done by airplane within expression:

$n = \dot n \cdot t$

If we know that $\dot n = 0.5\,\frac{rev}{min}$ and $t = 4\,min$, then we get that number of revolutions is:

$n = \left(0.5\,\frac{rev}{min} \right)\cdot (4\,min)$

$n = 2\,rev$

The airplane made two revolutions in four minutes.

Now, the distance covered by the airplane ($s$), measured in feet, is calculated by means of the equation below:

$s = 2\pi\cdot l\cdot n$

Where:

$l$ - Length of the string, measured in meters.

$n$ - Number of revolutions done by airplane, measured in revolutions (dimensionless).

If we get that $n = 2\,rev$ and $l = 4\,ft$, then the distance covered by the airplane in 4 minutes is:

$s = 2\pi\cdot (4\,ft)\cdot (2\,rev)$

$s \approx 50.265\,ft$

Airplane covers a distance of approximately 50.265 feet in 4 minutes.