1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

2.A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

3.An engineer is designing the runway for an airport. Of the planes which will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

## 7 comments:

sir...>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

kaupay man sa imo question>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>kalisod na kasayon

1.)given:a=3.20m/s2

t=32.8s

d=?

formula:d=1/2at2

solution:

d=1/2at2

=1/2(3.20m/s2)(32.8s)2

=1/2(3.20m/s2)(1075.848)

=1/2(344.688m)

d=1,721.344m or 1,721m...answer

2.)given:t=5.21s

d=110m

a=?

formula:

a=2d/t2

solution:

=2(110m)/(5.21s2)

=22om/27.1441s2

a=8m/s2...answer

3.)given:a=3m/s2

V=6s m/s

d=?

1st=?

formula:

1st=V/d

2nd d=1/2at2

solution:t=V/d=65m/s2/3m/s=225m/s..

d=1/2at2

=1/2(3m/s)(22s)2

=

1.)GIVEN:T=32.8 s

a=3.20 m/s2

d=?

fORMula:d=a/t

d=3.20 m/s/32.28 s

d=0.098 m is the answer...

2.)GIVEN:d=110 m

t=5.21 s

a=?

FORMULA:

a=2d/t2

SOLUTION:

=2(110m)/(5.21s2)

=22om/27.1441s2

a=8m/s is the answer

3.)GIVEN:V=6s m/s

a=3m/s2

d=?

1st=?

FOrmuLA:

1st=V/d

2nd d=1/2at2

sOLutION:t=V/d=65m/s2/3m/s=225m/s..

d=1/2at2

=1/2(3m/s)(22s)2

=66 is the ANswer... mary jeaN sERRATo

axa na gud ini sir....

1.)

GIVEN:

t=32.8 sec.

a=3.20m/s2

d=?

FORMULA:

d=Iv+1/2(a)t2

SOLUTION:

d=0+1/2(3.2m/s2)(32.8s*2)

d=1721.344m....answer

2.)

GIVEN:

t=5.21sec.

d=110

a=?

FORMULA:

a=2d/t2

SOLUTION:

a=2(110)/(5.21s2)

a=220m/27.1441s

a=8.1m/s2......answer

3.)

GIVEN:

a=3m/s2

V=65m/s

d=?

t=?

FORMULA:

d=1/2at2

t=V/a

SOLUTION:

t=(65m/s)/(3m/s2)

t=22sec........answer

d=1/2(3m/s2)(22s2)

d=726m.......answer

RIZA D> PRANSAL

1.)GIVEN:T=32.8 s

a=3.20 m/s2

d=?

fORMula:d=a/t

d=3.20 m/s/32.28 s

d=0.098 m is the answer...

2.)GIVEN:d=110 m

t=5.21 s

a=?

FORMULA:

a=2d/t2

SOLUTION:

=2(110m)/(5.21s2)

=22om/27.1441s2

a=8m/s is the answer

3.)GIVEN:V=6s m/s

a=3m/s2

d=?

1st=?

FOrmuLA:

1st=V/d

2nd d=1/2at2

sOLutION:t=V/d=65m/s2/3m/s=225m/s..

d=1/2at2

=1/2(3m/s)(22s)2

=66 is the ANswer... marlon ubaldo

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