Question

a plane starts from rest and accelerates in a straight line along the ground before takeoff. It moves 600m in 12s. Find its acceleration.

Answer 1

@amistre64

Answer 2

hi i know this is easy but i just kind of got a question about finding acceleration

Answer 3

so i know a=change in velocity/time

Answer 4

v = at

Answer 5

so the velocity would be 600/12 correct?

Answer 6

m/s = m/s^2 * s

Answer 7

seems fair to me

Answer 8

so i dont really understand why it wouldnt just be 50/12 for acceleration

Answer 9

but then if u do 50/12 then multiply that answer by 2 u get the right answer?

Answer 10

spose we have a constant accleration, in order to move we need to accelerate

position is: 1/2 at^2 + vo t + so

velocity is at + vo

acceleration is: a

Answer 11

lets use position to determine acceleration since that is what we are given

Answer 12

ok

Answer 13

are u saying vo is initial or final velocity?

Answer 14

600 = 1/2 a (12)^2

1200 = a (12)^2

a = 100/12

Answer 15

initial

Answer 16

and so is initial position, assumed to me 0 as well

Answer 17

that would be zero correct?

Answer 18

yes vo and so are 0 in this case

Answer 19

ok there are two more parts to this question, mind helping me out with them? u have been super helpful :D

Answer 20

sure ... if my memory holds up :)

Answer 21

ok sweet, the next part says find the speed at the end of 12seconds

Answer 22

any idea? :D

Answer 23

speed is just the velocity at t=12

Answer 24

since we know acceleration: a

v = at

Answer 25

so its just 8.3X12?

Answer 26

recall we are given a distance and time

d = 1/2 a t^2

600 = 1/2 a(12)^2

100/12 = 25/3 = a

Answer 27

v = 25(12)/3

Answer 28

why would u divide by 3 :o

Answer 29

i feel stupid for asking this stuff lol

Answer 30

100/12 reduces to 25/3

Answer 31

25*5/3*4 = 100/12

Answer 32

a = 25/3 and t=12

Answer 33

oh yea i thought 3 was like some other value or something

Answer 34

and the last part said find the distance moved during the 12th second

Answer 35

during the 12th second ... is that from t-12 to t=13?

Answer 36

*t=12

Answer 37

t=0 to 1 is the first second
so t=11 to 12 is the 12th second

Answer 38

yes i believe thats what it means but it just says the 12th second

Answer 39

600 - 1/2 a(11)^2

Answer 40

actually i think it might mean like 12th second to 13th second

Answer 41

your call, im going with 11 to 12 :)

Answer 42

0to1 =1
1to2 = 2
2to3 = 3

...

11to12 = 12

Answer 43

i have the answer but i need to figure out how to get to it so i guess we can just try one way

Answer 44

and then the other if it doesnt work

Answer 45

trying is the best way yes :)

Answer 46

so what u were doing was finding acceleration for the 11th second?

Answer 47

no finding the distance from t=11 to t=12 which is utilising the distance setup

1/2 a (12)^2 - 1/2 a (11)^2 = 1/2 a (12^2 - 11^2)

Answer 48

so u just use that equation u made?

Answer 49

lol, i didnt just 'make' it, but yes :)

Answer 50

i mean like the equation u just showed me how to get lol

Answer 51

if you know the distance at t=11 and the distance at t=12 ... then the difference between them is how far you hav emoved right?

Answer 52

yes

Answer 53

then lets go with 95.833333

Answer 54

ok sweet man !! u helped me so much!!!!

Answer 55

goodluck and all ;)