OK, got a test Monday, and my professor e-mailed us our review for siad test this past Friday. His tests are very similar to his reviews so did I do this right?(adsbygoogle = window.adsbygoogle || []).push({});

Question:

The motion of a particle is described as [itex]v=2x-\frac{9}{x} \ {\rm for}\ (0 \leq t \leq 10)s\ {\rm where}\ t_0=0,x_0=0[/itex]

Ok, this is what I did:

a) Determine the distance traveled in t=4 seconds

[tex]v=2x-\frac{9}{x}=\frac{dx}{dt}[/tex]

[tex]\frac{2x^2-9}{x}dx=dt[/tex]

[tex]\int \frac{2x^2-9}{x}dx=\int dt[/tex]

[tex]\int \frac{1}{4U}dU=t+c[/tex]

[tex]\frac{1}{4}\ln (2x^2-9)=t+c[/tex]

[tex]{\rm let }\ t=0,\ {\rm and }\ x=3[/tex]

[tex]c=\frac{ \ln (9)}{16}[/tex]

thus:

[tex]x^2=\frac{9}{2}e^{4t}+\frac{9}{2}[/tex]

[tex]x(4)\approx 6323.567 m[/tex]

b)How long will it take the particle to reach 0.02c?

[tex]0.02c \approx 1.8 \times 10^8[/tex]

so I plugged that into the original equation

[tex]1.8 \times 10^8 =\frac{2x^2-9}{x}[/tex]

[tex]2x^2-1.8 \times 10^8x-9=0[/tex]

[tex]x=\{ -5 \times 10^{-8},9.0 \times 10^7\}[/tex]

the first was a trivial solution so:

[tex]t=\frac{1}{4}\ln (2(9.0\times 10^7)^2-9)-\frac{\ln 9}{16}[/tex]

[tex]t=8.7816s[/tex]

c) Express acceleration as a function of x

[tex]a=v\frac{dv}{dx}[/tex]

[tex]a=(2x-\frac{9}{x})(2x-\frac{9}{x})\frac{d}{dx}[/tex]

[tex]a=(2x-\frac{9}{x})(2+\frac{9}{x^2})=4x-\frac{81}{x^3}[/tex]

d) Determine the acceleration at 0.02c

[tex]a=4(9.0\times 10^7)-\frac{81}{(9.0\times 10^7)^3}=3.6\times 10^8[/tex]

Is the above correct? I was talking to another student who did it differently and got a different answer. I want to make sure the above is correct so I don't miss this question on the test. Any insight would be greatly appreciated. Thanks.

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# Homework Help: Velocity and Acceleration Question

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