boomerang285
Please Help with Practice question!!!
An electrically charged particle accelerates uniformly from rest to speed v while traveling a distance x.
a) Show the acceleration of the particle is a= v^2/2x.
b) If the particle starts from rest and reaches a speed of 1.8 x 10^7 m/s over a distance of 0.10m, show that its acceleration is 1.6 x 10^15 m/s^2.
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boomerang285
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I know for B i just plug in the numbers...
shivaniits
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you know \[v ^{2}-u ^{2}=2as\]
and here s=x and u=0 since it started from rest so we have:-
\[v ^{2}-0=2ax\]
\[a=v ^{2}/2x\]
shivaniits
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now for second part just put the values for the equation..
hope this helps..!!
boomerang285
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so the equation for part A is ...
boomerang285
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v^2-0=2ax?
shivaniits
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yup that'w what you get after putting in the values x=s and u=0 in more general eqn \[v ^{2}-u ^{2}=2as\]
boomerang285
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ok, then for B it comes out like ...
boomerang285
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[1.8 x 10^7]^2/ {2*0.1) ?
shivaniits
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yes..
shivaniits
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hope this sorted you out..!!
boomerang285
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so it equals ...
boomerang285
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1.62e + 15 m/s^2 ?
boomerang285
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thats what comes up on a calculator.....
shivaniits
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oh..the answer is \[16.2 \times 10^{14}\]
boomerang285
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ok.......
boomerang285
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hmm wonder why mine is so off...
shivaniits
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everything alright..!!