Author Topic: Kinematics (Read 16746 times) Tweet Share

d0minicz · « **Reply #15 on:** July 05, 2009, 12:17:37 pm »

From a balloon ascending with a velocity of 10m/s a stone was dropped and reached the ground in 12 seconds. Given that the gravitational acceleration is 9.8 m/s , find:
a) the height of the balloon when the stone was dropped
b) the greatest height reached by the stone

thanks in advance...

Flaming_Arrow · « **Reply #16 on:** July 05, 2009, 12:24:00 pm »

a. v= - 10 a= 9.8 t= 12

solve for d for part

kamil9876 · « **Reply #17 on:** July 05, 2009, 12:36:05 pm »

yes to 2nd second question.

The stone goes up a bit, but is then slowed down by gravity until it's velocity is downwards.

a.) You should chose an appropriate positive y direction. If you decide to choose down as positive then $a=9.8$ , $u=-10$ , however if you choose up as positive then $a=-9.8, u=10$ . I'll do the latter:

$h=\frac{1}{2}at^2+ut$

sub in $a=-9.8, u=10, t=12$ .

h will be negative as it ends up on the negative side (what we chose as down). So the height is the negative of that.

b.) We want to know where the particle turns around, velocity equals 0 here.
$ v^2=u^2-2as$

v=0, u=10, a=-9.8.

Solve for s. This is that upward arrow distance he travels. (It is positive, as it's above the origin and we have chosen up to be positive,

d0minicz · « **Reply #18 on:** July 05, 2009, 12:47:27 pm »

What do i do once its solved for S ?
Also, couldnt you just let V = 0 (V= 9.8t - 10) and find the time at which it turns around and then sub t into the x equation
I dunno where to go from here.
thanks man

kamil9876 · « **Reply #19 on:** July 05, 2009, 12:51:06 pm »

Yes you could do that, it would also work.
Oh right and the distance s is the distance of that small upwards arrow, hence to get the total height above the ground you have to add the initial height of the balloon(ie that thing you worked out in part a)

d0minicz · « **Reply #20 on:** July 05, 2009, 12:54:22 pm »

ahhh thanks man

For this question: An object moves in a line with velocity given by $V = \frac {t}{1+t^2}$ . The object starts from the origin.
How would i work out the distance travelled in the 3rd second?

cheers again

kamil9876 · « **Reply #21 on:** July 05, 2009, 01:09:19 pm »

One thing you must be careful of is that the integral of velocity is displacement, not distance. As you may have already seen the magnitude of displacement is not neccesarily equal to the total distance travelled (like the balloon question, it's displacement was h but the total distance travelled was the sum of two arrows). Hence if there is a turning point you have to be careful, but if there is not turning point then its just simply the magnitude of displacement.

Looking at this equation, nowhere in the interval of time (2,3) is the velocity 0 or negative. Hence the object is moving in one direction. (the positive direction).

That means the distance is magnitude of displacement (change in position)

$x(t)=\int \frac{t}{1+t^2} dt $

and now just find $|x(3)-x(2)|$ .

A quick way woud be this:
$ |x(3)-x(2)|=|\int_2^3 x'(t) dt|$ (fundamental theorem of calculus)
$ |x(3)-x(2)|=|\int_2^3 v dt|$
$=|\int_2^3 \frac{t}{1+t^2} dt |$

And so the fact that it starts at the origin does not matter, distance travelled will always be the same for all values of c as the c's cancel out.

d0minicz · « **Reply #22 on:** July 05, 2009, 01:25:51 pm »

Hey can you please show me the other way to do it (instead of the fundamental calculus theorem) if you could
thanks

d0minicz · « **Reply #23 on:** July 05, 2009, 01:29:26 pm »

oops sorry dw about it mate got it
Hey so for finding the V = 0 within the (2,3) range ; as theres no T.P then theres no change for that range right?

kamil9876 · « **Reply #24 on:** July 05, 2009, 01:36:37 pm »

Fundamental theorem of calculus isn't as scary as the name suggests(It's pretty awesome IMHO), it's the thing u used in finding areas.
Other way:

$x(t)=\int \frac{t}{t^2+1} dt$

let u=t^2+1

\frac{du}{dt}=2t

$\therefore x(t)=0.5 \int \frac{1}{u} du$

$=0.5ln|1+t^2|+c$

You could find c here by subbing in t=0,x=0; but if you're clever you can just leave it and watch it cancel out.
$ x(3)-x(2)=0.5ln|1+3^2|+c - (0.5ln|1+2^2| + c) $

$=0.5(ln10 - ln4)$
$=0.5(ln\frac{5}{2}$

Quote

Hey so for finding the V = 0 within the (2,3) range ; as theres no T.P then theres no change for that range right?

WHat I meant was that because v never equals 0 when t is between 2 and 3, There is no turning point and hence we don't have to break it up but rather we can just find the change in position. because if you had a turning point you would have to sum up the distances travelling backwards and forwards. But here he is only travelling forwards so no need to worry.

d0minicz · « **Reply #25 on:** July 05, 2009, 01:39:53 pm »

ahh thanks alot !!!
also on the same question; How would you find the minimum acceleration?

kamil9876 · « **Reply #26 on:** July 05, 2009, 01:47:09 pm »

acceleration is $\frac{d}{dt}(v)$ . so differentiate velocity, you have acceleration. Now you want to find the minimum value of that function. Hence differentiate that function(acceleration function) and $\frac{d}{dt}(a)=0$ ... etc.

d0minicz · « **Reply #27 on:** July 05, 2009, 01:50:42 pm »

ah sweet cool
hmm im stuck with simplifying it
soo $a = \frac {1-t^2}{(1+t^2)^2}$
$\frac {da}{dt} = \frac {-2t(1+t^2)^2 - 4t(1+t^2)(1-t^2)}{(1+t^2)^4}$
dont know how to simplify =/
thanks in advance again sorry for all of the q's...

d0minicz · « **Reply #28 on:** July 05, 2009, 02:09:44 pm »

^^^^ Dw i got it

1. A body with constant acceleration starts with velocity 15 m/s. At the end of the eleventh second its velocity is 48 m/s. What is its acceleration?

I dunno what to do with the "eleventh second" thing. Is it (10,11) ?

2. A car accelerates uniformly from 5km/h to 41 km/h in 10 seconds. Express this acceleration in:
a) $km/h^2$
b) $m/s^2$

thanks again

zzdfa · « **Reply #29 on:** July 05, 2009, 02:16:12 pm »

ask yourself;
whats the change in velocity?
whats the change in time? i.e. over what time period did the velocity change?

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