# AN AIRPLANE ACCELERATES FROM REST ON A RUNWAY AT 5.50 M/S FOR 20.25 S UNTIL AT FINALLY TAKES OFF .WHAT IS THE DISTANCE COVERED BEFORE TAKE OFF?

1
by aljoseph

2015-02-12T18:28:45+08:00

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This is kind of confusing because you haven't stated your acceleration but let me try.

If the plane started at rest (which would be 0 m/s) and flies up when it reaches 5.5 m/s 20.25 seconds later then:

First, find acceleration:

a= vf-vi/t

a= 5.5m/s / 20.25 s
a = 0.272 m/s2

Then find the distance:

d = vit + at2

d = (0m/s x 20.25 s) + (0.272 m/s2 x (20.25 s)2)
d = 0 + (0.272 m/s2 x 410.0625 s2)
d = 55.69 m

Correction: I forgot to write 1/2 in the formula d = vit + at2. It should be d = vit + 1/2at2. However, the answer is still correct.
mali naman
botogss
sorry pero pwede ko bang malaman kung paano kinhuha yung tamang sagot. Ang problem mo kasi ay parang uniform accelerated motion. So ang ginawa ko ay hinanap ko muna ang acceleration, tsaka ko kinuha ang distance. Please.