Mechanics Q (1 Viewer)

cutemouse · Jun 8, 2009

Hello, could someone please do this for me? Thanks

A particle moves in a straight line with retardation which increases uniformly with the distance moved. Initially the retardation is 5ms^-2 and when the particle has moved a distance of 12m the retardation is 11ms^-2. Find the distance moved by the particle in coming to rest if the initial velocoity is 20ms^-1

Thanks

jet · Jun 8, 2009

I am not sure how correct i am.
$\text{If the retardation increases uniformly, then it is linear, i.e} \ddot{x} = ax + b \\ x = 0, \ddot{x} = -5, b = -5 \\ x = 12, \ddot{x} = -11, 12a - 5 = -11, a = -\frac{1}{2}\\ \ddot{x} = -\frac{x}{2} - 5 \\ \frac{d}{dx}\left (\frac{1}{2}v^2 \right ) = -\frac{x}{2} - 5 \\ \frac{1}{2}v^2 = -\frac{1}{4}x^2 - 5x + C \\ x = 0, v = 20, C = 200 \\ \therefore \frac{1}{2}v^2 = 200 - 5x - \frac{1}{4}x^2 \\ \text{When } v = 0, x^2 + 20x -800 = 0 \\ (x + 40)(x - 20) = 0 \\ \therefore x = 20$

study-freak · Jun 8, 2009

$\\a=-kx+c=\frac{1}{2} \frac{dv^2}{dx} \\when x=0, a=-5, v=20 \\-5=c \\\therefore a=-kx-5 \\when x=12, a=-11 \\-11=-12k-5 \\-k=-\frac{1}{2} \\\therefore a=-\frac{1}{2}x-5 \\v^2=-\frac{1}{2}x^2-10x+c \\when x=0, a=-5, v=20 \\400=c \\\therefore v^2=-\frac{1}{2}x^2-10x+400 \\when v=0, \\0=-\frac{1}{2}x^2-10x+400 \\0=x^2+20x-800 \\(x+40)(x-20)=0 \\x=-40 or 20 \\but x\geq 0 \\\therefore x=20m$
Got beaten but I did it anyway so I'll post it lol...

cutemouse · Jun 8, 2009

Thanks alot jetblack, now I understand why the solutions said that accel = c-kx

(I understood the kx bit, but no clue why it had a c, but now I get it =D)

Mechanics Q (1 Viewer)

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