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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Chavi on July 28, 2010, 05:28:38 pm

Title: Kinematics help
Post by: Chavi on July 28, 2010, 05:28:38 pm: Hey forum,

I'm wondering if I can get some help with the following kinematics question:

1. A particle is brought to top speed with an accel which varies linearly as the distance travelled. It starts from rest with an accel of 3 m/s^2 and reaches top speed in a distance of 160m. Find:
a) the top speed
b) the speed when the particle has moved 80m

Thanks.
Title: Re: Kinematics help
Post by: 98.40_for_sure on July 28, 2010, 05:56:04 pm: Do you have the answers? I need something to check with, before i post up how i did it
Title: Re: Kinematics help
Post by: Martoman on July 28, 2010, 08:35:40 pm: First one should be $8\sqrt{15}$

Second $8\sqrt{90} = 24\sqrt{10}$ :P
Title: Re: Kinematics help
Post by: Chavi on July 28, 2010, 08:37:03 pm: Quote from: Martoman on July 28, 2010, 08:35:40 pm
First one should be $8\sqrt{15}$

Second $8\sqrt{90} = 24\sqrt{10}$ :P

According to the answers
a) 4sqrt(30) m/s
b) 6sqrt(10) m/s
Title: Re: Kinematics help
Post by: Martoman on July 28, 2010, 08:53:30 pm: The way I got to the first one was:

let T be some time in the future at top speed x.

$.5Tx = 160$ ie: a pretty triangle.

$x = \frac{320}{T}$

Also gradient of line is 3.

$3 = \frac{x-0}{T-0}$

$3T = x \Rightarrow 3T^2 = 320$

$T = 10.33$

So x, top speed, = 320/10.33 = 30.98... $8\sqrt{15}$

this assumes constant acceleration RTFQ
Title: Re: Kinematics help
Post by: Chavi on July 28, 2010, 09:02:50 pm: my workings thus far have been:
when $x=0, a=3$
so $a=kx+3$
$a=kx+3$

$\frac{d}{dx}(\frac{v^2}{2})=kx+3$
so find that c =0, and:
$v^2=kx^2+6x$

now, how do you find k?
Title: Re: Kinematics help
Post by: Martoman on July 28, 2010, 09:16:15 pm: Ok.

Simply you have $a = kx + 3$

When x = 160, a = 0 as it accelerates no more!

$k = \frac{-3}{160}$

Shove it in it works out to your answer for 1.

Then you have your v equation.

Shove in 80 for part 2.

Works out nicely.
Title: Re: Kinematics help
Post by: Chavi on July 28, 2010, 09:21:25 pm: Quote from: Martoman on July 28, 2010, 09:16:15 pm
Ok.

Simply you have $a = kx + 3$

When x = 160, a = 0 as it accelerates no more!

$k = \frac{-3}{160}$

Shove it in it works out to your answer for 1.

Then you have your v equation.

Shove in 80 for part 2.

Works out nicely.
Thanks - i was experimenting with diff values - just overlooked the correct figure.