Question

Suppose that the distance an aircraft travels along a runway before takeoff is given by Upper D equals (5 divided by 3 )t squaredD=(5/3)t2​, where D is measured in meters from 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 300300 km divided by hkm/h. How long will it take to become​ airborne, and what distance will it travel in that​ time?

Answer 1

a. Time=25seconds

b.distance=1041.67m

Step-by-step explanation:

a.The equation for
`D` in terms of m/s is
`(250)/(3)m/s` after conversion.

To find when speed reaches 300km/hr=83.33m/s, we find
`D\prime` and solve for
`t`

`D=(5)/(3)t^2\\D\prime=(5)/(3)(2t)=(10)/(3)t=(250)/(3)\\t=25sec`

b. From a, above we already have our t=25seconds as the time it takes before the plane is airborne.

#To find distance travelled in that time , we substitute for
`t=25` in our distance equation:

`D(25)=(5)/(3)(25)^2\\=1041.67m`

Hence the distance of the plane before it gets airborne is 1041.67m