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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Tea.bag on August 19, 2008, 09:45:41 pm

Title: Kinematics
Post by: Tea.bag on August 19, 2008, 09:45:41 pm: i cant do this question!!
can someone explain how to do this??

1. a body at rest starts moving with an initial acceleration of 5m/s^2. the acceleration decreases uniformly with the distance travelled reaching a value of zero when the body has travelled 80 metres. Find the maximum speed of the body.
Title: Re: Kinematics
Post by: Collin Li on August 19, 2008, 09:53:47 pm: Acceleration decreases uniformly, hence acceleration is a linear function of distance travelled:

Since the body starts at rest, displacement is distance (as there is no change in direction, and acceleration is only decreasing -- no deceleration):

$a = mx + c$

You know that (0,5) initially, and (80, 0) when the acceleration decreases to zero after 80 metres.

Hence: $a = -\frac{1}{16}x + 5$

To find maximum speed of body, we want to find velocity. In order to get a velocity, we must use the fact that: $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$ (the other ones will not work - they do not have the right variables).

Therefore: $\frac{d}{dx}\left(\frac{1}{2}v^2\right) = -\frac{1}{16}x + 5$

$\implies \frac{1}{2}v^2 = -\frac{1}{32}x^2 + 5x + C$

Since the body starts at rest, (0,0), $\implies C = 0$

$\implies \frac{1}{2}v^2 = -\frac{1}{32}x^2 + 5x$

The body stops accelerating at $x=80$ , so this will be its maximum speed:

$\frac{1}{2}v^2 = -\frac{80^2}{32} + 5(80)$

$\implies v = \sqrt{400} = 20 \mbox{ ms}^{-1}$
Title: Re: Kinematics
Post by: Mao on August 19, 2008, 09:55:10 pm: $\frac{da}{dx}=-k$

$\implies a=-kx+a_0$

   $a_0=5, k=\frac{5}{80}=\frac{1}{16}$

$\implies \frac{d}{dx}\left(\frac{1}{2}v^2\right)=-\frac{x}{16}+5$

$\implies \frac{1}{2}v^2=-\frac{x^2}{32}+5x+C$

   given that when x=0, v=0, $\implies C=0$

$\implies \frac{1}{2}v^2=-\frac{x^2}{32}+5x$

   since acceleration was positive all the way until displacement is 80, the maximum velocity is reached at x=80

$\implies \frac{1}{2}(v_{max})^2=-\frac{80^2}{32}+400$

$\implies v_{max}=20$

EDIT: oops arithmetic error at last step
Title: Re: Kinematics
Post by: Tea.bag on August 19, 2008, 09:58:08 pm: two diffrent answers??
which one is right
btw my book says 20m/s so i dont know if the book is right or not
Title: Re: Kinematics
Post by: Collin Li on August 19, 2008, 10:00:11 pm: My one did the linear equation incorrect. I will try to fix it.
Title: Re: Kinematics
Post by: Mao on August 19, 2008, 10:02:00 pm: sorry, it is 20. i made an arithmetic error at the last step
Title: Re: Kinematics
Post by: Tea.bag on August 19, 2008, 10:07:45 pm: thnx a lot guys...
this whole time i was trying to find velocity in terms of time..i over looked the fact that acceleration can be in terms of position :D
Title: Re: Kinematics
Post by: Tea.bag on August 26, 2008, 09:55:46 pm: another question.

1) A car slows down with a constant retardation from 24m/s to 16m/s over a distance of 15 metres. What further distance will it travel before coming to rest?
Title: Re: Kinematics
Post by: Flaming_Arrow on August 26, 2008, 10:05:15 pm: Quote from: Tea.bag on August 26, 2008, 09:55:46 pm
another question.

1) A car slows down with a constant retardation from 24m/s to 16m/s over a distance of 15 metres. What further distance will it travel before coming to rest?

$v = 16 u =24 d=15 a= ?$

$v^2 = u^2 + 2ad$

$256 = 576+ 2*a*15$

$-320 = 30a$

$a = -10.7 ms^{-2}$

$a = -10.7 v = 0 u =16 d= ?$

$v^2 = u^2 + 2ad$

$0 = 256 + 2 * -10.7 * d$

$-256 = -21.4d$

$d = 11.96m$

hence it will travel 12m before it comes to a rest

duno if this is right
Title: Re: Kinematics
Post by: shinny on August 26, 2008, 10:06:52 pm: Its...mostly correct. Just dont round off the acceleration (its $-\frac{32}{3}$ ) and you'll get 12 metres exactly.
Title: Re: Kinematics
Post by: Tea.bag on August 26, 2008, 10:10:40 pm: thnx..

1 more :P

Two trains pass one another, travelling in opposite directions on parallel tracks. when the fronts of the trains are in line they are travelling at 12m/s and 16m/s and accelerating at 0.5m/s^2 and 1m/s^2 respectively. the length of each train is 136 metres. how long does it take the rear of the two trains to pass?
Title: Re: Kinematics
Post by: /0 on August 26, 2008, 10:16:03 pm: The relative velocity of one train with respect to the other is $12+16 = 28 m/s$ by vector addition. Likewise, as acceleration is a vector, you can see that the acceleration of one train with respect to the other is $1.5m/s^2$

Hence you can use

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

$136 = 28t + 1.5t^2$

$t = 4$ seconds. ( i think )

(yeah, oops, the distance should be twice 136m = 272 m) but hey, that still gives the wrong answer :/ t = 7
Title: Re: Kinematics
Post by: Tea.bag on August 26, 2008, 10:17:37 pm: nope the answer says 8 seconds
Title: Re: Kinematics
Post by: Flaming_Arrow on August 26, 2008, 10:21:18 pm: Quote from: DivideBy0 on August 26, 2008, 10:16:03 pm
The relative velocity of one train with respect to the other is $12+16 = 28 m/s$ by vector addition. Likewise, as acceleration is a vector, you can see that the acceleration of one train with respect to the other is $1.5m/s^2$

Hence you can use

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

$272 = 28t + 1.5t^2$

$t = 4$ seconds. ( i think )

(yeah, oops, the distance should be twice 136m = 272 m) but hey, that still gives the wrong answer :/

u forgot to divide acceleration by 2

$272 = 28t + 0.75t{^2}$

$t = 8s$
Title: Re: Kinematics
Post by: Tea.bag on August 26, 2008, 10:25:09 pm: why do u add the velocities and the acceleration together??
Title: Re: Kinematics
Post by: xox.happy1.xox on August 26, 2008, 10:29:56 pm: You add the velocities and acceleration because you want to find out when the rears are just touching each other. I always like to assume that one is stationary, therefore, for them to actually touch rears, the velocities and accelerations need to be added together. I'm sure there is a better explanation :P
Title: Re: Kinematics
Post by: shinny on August 26, 2008, 10:33:24 pm: Its because theyre moving in opposite directions. So if you define the direction system so that the positive direction for each separate system (each train basically) as its direction of movement, then you'll end up with two positive accelerations/velocities.

EDIT: typo