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1. I am stuck on this question from M1, since we have been doing dynamics and did SUVAT (equations of motion) stuff ages ago, I've completely stumped now on how to do this question.

In taking off, an aircraft moves on a straight runway AB of length 1.2 km. The aircraft moves from A with intial speed 2 m/s. It moves with constant acceleration and 20s later it leaves the runway at C with speed 74 m/s. Find:

(a) the acceleration of the aircraft,
(b) the distance BC.

2. ok lets do a) first,

s = x (we dont have length AC) , u = 2, v= 74, a=?, t = 20

What equation can we use now?
3. this is a suvat equation !

s = ut + 1/2 at^2
4. a) v=u+at
b) Find AC
5. (Original post by Eaglepowers)
this is a suvat equation !

s = ut + 1/2 at^2
You don't know s or a.
6. you can use V = U + AT to calculate acceleration and with that calculate the distance of BC.

so the acceleration would be: 3.6.

and the distance of AC would be 2612m, thus BC would be 2612 - 1200 = 1412m therefore, 1.412km ?
you can use V = U + AT to calculate acceleration and with that calculate the distance of BC.

so the acceleration would be: 3.6.

and the distance of AC would be 2612m, thus BC would be 2612 - 1200 = 1412m therefore, 1.412km ?
Your acceleration is right but AC can't be greater than AB.

AC = ut + 1/2 at^2

and BC = AB - AC
8. a) v=u+at

this should give a value of a = 3.6 m/s/s

b) Then work out length AC S=0.5(u+v)t i got this to be 760m

Subtract this from the length of AB 1200 - 760 = 440m

(note i did this in my head. You may want to check first. Hope it helped)
9. For those confused. The point C lies inbetween points A and B

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