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The distance an aircraft travels along a runway before takeoff is given by $D=(10/9)t^2$, where $D$ is measured in meters form the starting point and $t$ is measured in seconds from the time the brakes are released.

The aircraft will become airborne when its speed reaches $400\;\text{km/h}$.

$\text{(1)}$ How long will it take to become airborne, and
$\text{(2)}$ What distance will it travel in that time?

My try:

Finding the derivative of $D$ will give the velocity function $v(t)$.
Equate $v(t)$ with $400\;\text{km/h}$ to find $\text{(1)}$:

$\frac{\text{d}}{\text{dt}}D=\frac{\text{d}}{\text{dt}}(10/9)t^2 = v(t)=(20/9)t$

$400=(20/9)t$

$(\frac{9}{20})(400)=t$

$180\; \text{sec}=t$

This is wrong though. The correct answer is $50\; \text{sec}=t$ for $\text{(1)}$. Thank you.

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Notice that $t$ is measured in seconds and $D$ in metres meaning that $\tfrac{d}{dt}D$ will be is m/s.

The take-off speed if given in km/h. So that's where your problem comes from.

Notice that 1 km = 1,000 m and 1 h = 3600 s. It follow that

$$1 \, \text{km/h} = \frac{1 \, \text{km}}{1 \, \text{h}} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s}$$

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