Question

Suppose that the distance an aircraft travels along a runway before takeoff is given by D=(5/3)t^2, where D is measured in meters from the starting point and it measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 300km/h. How long will it take to become airborne, and what distance will it travel in that time?

Answer 1

Answer:

It will take 90s for the aircraft to become airbone.

In that time, the aircraft will have traveled 13500m = 13.5km.

Step-by-step explanation:

The velocity function is the derivative of the distance function.

In this problem, we have the following distance function:

$D(t) = \frac{5t^{2}}{3}$

So the speed function is the following:

$S(t) = \frac{10t}{3}$

The aircraft will become airborne when its speed reaches 300km/h. How long will it take to become airborne?

This is going to happen when S(t) = 300. So:

$S(t) = \frac{10t}{3}$

$\frac{10t}{3} = 300$

$t = 90$

It will take 90s for the aircraft to become airbone.

What distance will it travel in that time?

This is D(t) when t = 90. So:

$D(t) = \frac{5t^{2}}{3}$

$D(90) = \frac{5*90^{2}}{3} = 13500$

In that time, the aircraft will have traveled 13500m = 13.5km.