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?

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:

So the speed function is the following:

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:

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:

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

(c, 0) i might be wrong so don't put full faith on my answer

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )