first ten correct answers gets the points. Solve the problem below.

An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 4.6 m/s^{2}. The takeoff speed for this plane will be 80 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

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d = 695.65 m.

yes. you’re first.

695.65m

2nd right.

695.7 meters

3rd right.

695.652 meters – minimum allowed length for the runway

4th right.

695.65 meters or 696 meters is the minimum allowed length for the runway.

5th right.

695.65 m

6th right.

695.66 m ,minimum allowed length

7th right.

amo na ba maam?

yup.

d= 695.65m (length of the runway)

you’re 8th.

the minimum allowed length is 695.65 meters.

9th.

the answer is 695.65 meters.

10 th. congrats you’re the cutoff.

the minimum length allowed for the runway is 695.65 meters.

the answer is 695.65 meters mam.but i’m not sure.

Solution:

vf2(squared)=vi2(squared) + 2ad

(80m/s)raise to 2=(o)+2(4.6m/s squared)d

6400m2/s2=9.2m/s2 *d

so that you can get the (d)distance,over the 9.2m/s2 on both sides.

and now you have 6400m2/s2/(9.2m/s2)=d or 6400 m2/s2 divided by 9.2 m/s2=d

(ma’am sorry I don’t know how to raise an exponent in the computer,I’ll just write it as m/s2..sorry talaga ma’am.peace ^-^ 🙂

the answer is 695.65217 m. or 695.65 m

🙂

695.65 m is the minimum allowed length for the runway.

the minimum allowed length for the runway is 695.65 m/s :)))

unit should be meter.

vf² = vi² + 2 a d

0 = (80 m/s)² + 2 (4.6 m/s²) d

0 = 6400 m²/s² + (9.2 m/s²) d

9.2 m/s² d = 6400 m²/s²

———- = ———–

9.2 m/s² d = 9.2 m/s²

d = 695.65 m ← answer

amo naba ????

695.65 m…