The problem is as follows:

The given alternatives in my book are:

$\begin{array}{ll}

1.&4\,\frac{m}{s}\\

2.&8\,\frac{m}{s}\\

3.&16\,\frac{m}{s}\\

4.&20\,\frac{m}{s}\\

\end{array}$

I'm not sure if my attempt is correct in the solution of this problem but I thought that to get the average speed is given by:

$\overline{v}=\frac{s}{t}$

Since it mentions that it is a rotation with constant angular acceleration, then what I need is the acceleration. Using the information from the picture I'm getting this:

Considering $r=2\,m$

$\omega_{1}=\frac{v}{r}=\frac{10}{2}=5$

$\omega_{2}=\frac{v}{r}=\frac{14}{2}=7$

Therefore the acceleration can be found:

$\omega_{2}^2=\omega_{1}^2+2\alpha\Delta\theta$

$7^2=5^2+2\alpha\left(3\right)$

Therefore:

$\alpha=\frac{49-25}{6}=4\,\frac{rad}{s^2}$

Since what they request is the average speed then what is needed is the displacement for the given speeds.

$10^2=6^2+2\left(4\right)\Delta\theta$

$\Delta\theta=\frac{100-36}{8}=8\text{rad}$

But the time elapsed for that displacement is also required for getting the average speed:

Then:

$\omega_{f}=\omega_{o}+\alpha t$

$10=6+\left(4\right)t$

$t=1$

So it is only $1\,s$ elapsed in the given interval.

Therefore the average speed would be:

$\overline{v}=\frac{s}{t}=\frac{8\times 2}{1}=16\,\frac{m}{s}$

Which is what appears in alternative number $3$. But is my solution correct?.

At an air exhibition a biplane is flying in circles as shown in the figure from below. It is known that type of motion of this airplane is a rotation with constant angular acceleration. The radius of the circle is $2\,m$. Using this information find the average speed in $\frac{m}{s}$ of the plane between the instants when $\omega=6\,\frac{rad}{s}$ and $\omega=10\,\frac{rad}{s}$.

The given alternatives in my book are:

$\begin{array}{ll}

1.&4\,\frac{m}{s}\\

2.&8\,\frac{m}{s}\\

3.&16\,\frac{m}{s}\\

4.&20\,\frac{m}{s}\\

\end{array}$

I'm not sure if my attempt is correct in the solution of this problem but I thought that to get the average speed is given by:

$\overline{v}=\frac{s}{t}$

Since it mentions that it is a rotation with constant angular acceleration, then what I need is the acceleration. Using the information from the picture I'm getting this:

Considering $r=2\,m$

$\omega_{1}=\frac{v}{r}=\frac{10}{2}=5$

$\omega_{2}=\frac{v}{r}=\frac{14}{2}=7$

Therefore the acceleration can be found:

$\omega_{2}^2=\omega_{1}^2+2\alpha\Delta\theta$

$7^2=5^2+2\alpha\left(3\right)$

Therefore:

$\alpha=\frac{49-25}{6}=4\,\frac{rad}{s^2}$

Since what they request is the average speed then what is needed is the displacement for the given speeds.

$10^2=6^2+2\left(4\right)\Delta\theta$

$\Delta\theta=\frac{100-36}{8}=8\text{rad}$

But the time elapsed for that displacement is also required for getting the average speed:

Then:

$\omega_{f}=\omega_{o}+\alpha t$

$10=6+\left(4\right)t$

$t=1$

So it is only $1\,s$ elapsed in the given interval.

Therefore the average speed would be:

$\overline{v}=\frac{s}{t}=\frac{8\times 2}{1}=16\,\frac{m}{s}$

Which is what appears in alternative number $3$. But is my solution correct?.

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