Dekoyl's Questions

[kamil9876]

Q:A vehicle with constant acceleration travels 20m in 2s and 13m in the next second. What is the acceleration?

Thanks =\

let u= initial velocity, a =acceleration

since 3 seconds total:

    -----1


                   -----2

       -----1
    -----2

solve simultaneous equations for u gives

    ,

Nah, I think by "travelling 13m in next second" it means that the distance covered in the interval (2,3) is 13m. Hence the overall position from the origin at t=3 is 20+13 (assuming direction has not changed).

Oh and just realising, I think u misentepreted distance as velocity.

[kamil9876]

Are you saying that:

average velocity=instantaenous velocity?

[Damo17]

Are you saying that:

average velocity=instantaenous velocity?

LOL, I'm so stupid. Get it now.

[Flaming_Arrow]

just wondering, why are you doing unit 3 stuff? or is this from spesh?

[kamil9876]

Surprisingly you still get the same acceleration. I think doing this exact same incorrect method woudl still give the same acceleration because of funny relationships in constant acceleration:

ie:

Is true only in constant acceleration, in non-constant acceleration this isn't true. Hence the reason why the average velocity 13 is halfway in between 12 and 14. I think you can prove using this that this exact incorrect method gives the correct acceleration, buit only in constant acceleration.

[dekoyl]

Q: A lift descends a distance of 30m, from rest to rest, as follows: constant acceleration for the first 10 m, constant velocity for he next 10 m, constant deceleration for the next 10 m. If the total time taken is 5 s, find the greatest speed reached by the lift.

Thanks =]

[TrueTears]

...[1]

...[2]

x = 10 a = 1

Therefore greatest speed reached is

[dekoyl]

Q: A particle starting from point O with uniform velocity 4 m/s. 2 seconds later, another particle leaves O in the same direction with a velocity of 5 m/s and with an acceleration of . Find when and where it will overtake the first particle.

I know there's something wrong with my understanding but what I had constructed so far was this:

Red: First particle -
Blue: Second particle - <--- I suspect this is the wrong.

Thanks for the help. I feel very slow in these holidays =(

[kamil9876]

Diagram way:

you want to find a line t=T such that the area "bounded by the line t=T, the blue line, the line t=2 and the t axis" equals the area bounded by "t=0, t=, the red line, and the t axis".

Reason: They start at same point hence at time T they should have the same displacement hence same area under graph.



Find equations of blue and red and integrate and find T. There are hundreds of way of picturing this, this is just what I came up with now.

[kamil9876]

Blue: Second particle - <--- I suspect this is the wrong.

Thanks for the help. I feel very slow in these holidays =(

ie: for that particle since it travels two seconds less.

[dekoyl]

Thanks kamil.
For the 'blue' equation, I obtained it from
so I didn't think that the travelling two seconds less needed to be taken into account.

For some reason I didn't get it after integrating but I'll keep trying.

[kamil9876]

No it doesn't, the first post i offered my solution, in the second post I fixed up your attempt. You should think of them seperately, I think you confused the two.

[kamil9876]

here is the blue one:






So you see it is equivalent to my second post

[dekoyl]

^Ah yeah. Thanks for that, Kamil.

Anyway I think I've given up on this question. It's either the answers are wrong (it says T=2 seconds, with distance of 16) but I keep getting T=4. I used your approach and another approach and I keep getting T=4 seconds.

It's just the answers to Fitz is rarely wrong so I'm doubting myself at the moment

[dekoyl]

Hopefully the last one on velocity/time graphs... I don't know why but I seem to have quite a bit of trouble with these =\

A train passed a station, A, at 60 km/h, maintained this speed for 12km, and was then uniformly retarded to stop at B, 15 km from A. A second train started from A the instant the first train passed it, was uniformly accelerated, then kept at constant speed over 3 km, and finally retarded to stop at B at the same time as the first train. Find the constant speed of the second train.

I found the retardation of the first train (I think). , time kept at constant speed: 720 seconds, and total time retarded (360 seconds). but I'm not sure how I could use this info for the second train.

Thanks!