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

Answer:

a. Time=25seconds

b.distance=1041.67m

Explanation:

a.The equation for $D$ in terms of m/s is $\frac{250}{3}m/s$ after conversion.

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

$D=\frac{5}{3}t^2\\D\prime=\frac{5}{3}(2t)=\frac{10}{3}t=\frac{250}{3}\\t=25sec$

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:

$D(25)=\frac{5}{3}(25)^2\\=1041.67m$

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