Can anyone explain this to me: "An electron is accelerated from rest to 3.0 X 10^6 m/s in 5.0 X 10^-8 s. What distance does the electron travel in this time interval?" The book says 75mm, but i got 150mm. Any idea why i need to divide my answer by 2???? Please help!

Question

amistre64 · Answer

what formulas do you have to work with?

amistre64 · Answer

we can work them with some calculus if need be

anonymous · Answer

For formulas i was simply multiplying the two together so as to cancelk out the s

anonymous · Answer

If there's a calculus formula, that would be great..... this is from a calc based physics class

anonymous · Answer

You probably derived it incorrectly. Did you end with d=vt where v is the final velocity?

anonymous · Answer

I don't think you need calc for this.

amistre64 · Answer

if we assume a constant acceleration,  then a(t) = k

we are given a velocity that it reaches in a certain time limit, a boundary value.

v(t) is the integration of a(t):  v(t) = kt + c

anonymous · Answer

I dont think calc is really needed either. And as far as the question from the book, thats it word for word

amistre64 · Answer

distance is the integration of velocity giving us:  s(t) = kt^2/2 + ct + d

amistre64 · Answer

since the intial velocity is at rest, c=0 when t=0 and we can determine k

amistre64 · Answer

and d=0 when t=0 since we are at the start of it ...

amistre64 · Answer

calc isnt needed if you have the formulas given to you already

anonymous · Answer

i kinda just thought this was a simple conversion problem....

amistre64 · Answer

but i work better with the calculus way :)

amistre64 · Answer

v = 3.0 X 10^6 m/s in t = 5.0 X 10^-8 s

k = v/t

s(t) = kt^2/2

amistre64 · Answer

so:  vt^2/2t = vt/2

amistre64 · Answer

which is why its divided by 2 lol

anonymous · Answer

For the non-calc way, |dw:1410719739067:dw|
Solve the second equation for a and then plug and chug.

(Just in case you want this)