integral prob: particle velocity v = 0.1564x; find accel.

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ku1005

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Oct 27, 2006
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Velocity of a particle moving along the x-axis is given by v = 0.1564x , where v is in mm/s and x is in mm.

If x = 6.2mm whent t = 0 determine the value of acceleration when t = 7

I said we know

x = 6.2 mm when t = 0 and from v = 0.1564x = 0.96968mm/s at t = 0

we want

x = ??mm when t = 7 which we can get from v = ....at t = 7

so using fact that

v dt = dx

0.1564s dt = dx

Integrate

(0 -> 7) 0.1564 dt = (6.2->s)1/x dx

ln s = 1.0948 + ln 6.2

thus s = 18.529mm when t = 7

NOW

v must = 2.8929mm/s from v formula ( v = 0.1564x)

thus using

a dt = dv

ie integrate

(0->7)a dt = (0.9697->2.8929) dv

7a = 1.9283

a = 0.27547

WHICH IS INCORRECT??

but why??

any tips on mky working wuld be great!
 
ku1005 said:
Velocity of a particle moving along the x-axis is given by v = 0.1564x , where v is in mm/s and x is in mm.
Click to expand...
Shouldn't x be time instead? If x is mm and the integral is taken, then the next equation wouldn't make sense. And you didn't include t in your first equation. So how can you set t to 7? I think you may have read the problem wrong...
 
you were given velocity as a function of position ...

v = kx

dx/dt = kx

dx/x = k dt

ln|x| = kt + C

x = Ae[sup:378n3eaf]kt[/sup:378n3eaf], where k = 0.1564

at t = 0 sec, x = 6.2 mm ...

6.2 = Ae[sup:378n3eaf]0[/sup:378n3eaf] , therefore A = 6.2

x = Ae[sup:378n3eaf]kt[/sup:378n3eaf], where k = 0.1564 and A = 6.2

v = dx/dt = Ake[sup:378n3eaf]kt[/sup:378n3eaf]

a = dv/dt = Ak[sup:378n3eaf]2[/sup:378n3eaf]e[sup:378n3eaf]kt[/sup:378n3eaf]

a(7) = 0.4532 mm/s[sup:378n3eaf]2[/sup:378n3eaf]
 
Thanks Skeeter, I will try that, it follows a similar rationale to mine, except im still not sure why my method gave a differnt answer, im pretty sure its because you keep yours expressed as a function, whereas in mine I choose certain values, neverless, im appreciative of your input!
 
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