anonymous
  • anonymous
An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 ms. What is the magnitude of this acceleration?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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JamesJ
  • JamesJ
What's the equation of motion for a body under constant acceleration?
JamesJ
  • JamesJ
Suppose v = initial velocity t = time a = constant acceleration Then what is the formula for the distance, d, the object moves in time t? d = ... what?
Kainui
  • Kainui
What is the definition of acceleration? It's the change of velocity with respect to time. How would you put this in the form of an equation?

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JamesJ
  • JamesJ
I'm going to change notation slightly, writing now u for initial velocity. Then we have: u = initial velocity t = time a = constant acceleration Write also v(t) = velocity of object at time t d(t) = position of object at time t Now, by definition, acceleration = (change in velocity)/(Time) hence \[ a = \frac{v(t) - u}{t} \] thus \[ v(t) - u = at \] and \[ v(t) = u + at \] Making sense so far?
anonymous
  • anonymous
Yeah, I'm following
JamesJ
  • JamesJ
Now, what's d(t) ?
anonymous
  • anonymous
The position?
JamesJ
  • JamesJ
yes, what's the formula for it?
anonymous
  • anonymous
v1t + 1/2 a t^2?
JamesJ
  • JamesJ
Yes \[ d(t) = ut + \frac{1}{2}at^2 \] For your problem, you're told that the initial velocity is zero, t = 0.5 ms and d(0.5 ms) = 0.92 m Now calculate a.
JamesJ
  • JamesJ
*correction, d(0.5 ms) = 0.02 m, not 0.92.
anonymous
  • anonymous
would I try and get a by its self?
JamesJ
  • JamesJ
Thus u = 0 m/s t = 0.005 s d = 0.02 m Substitute that into the equation above and solve for a
anonymous
  • anonymous
.02 = 0(.005) + .5(a)(.005)^2?
JamesJ
  • JamesJ
yes
anonymous
  • anonymous
So if a=(v(t)−u)/t, that would equal v(t)/t, correct?
JamesJ
  • JamesJ
yes
JamesJ
  • JamesJ
but here velocity is not constant so you have to be very careful with that formula
anonymous
  • anonymous
But the velocity equation is dependent on the acceleration equation: v(t)=u+at
JamesJ
  • JamesJ
Yes, so here because acceleration is constant, we're good. I.e., the average acceleration is equal to the constant acceleration.
anonymous
  • anonymous
But the closest we got to solving for a is: .02 = 0(.005) + .5(a)(.005)^2 and v(t) needs a, v(t)=u+at I'm not getting this part, how do we solve for either one?
anonymous
  • anonymous
?
JamesJ
  • JamesJ
The question only asks you to find the acceleration. Hence you only need one equation and the one that is germane here--because it uses the information of the problem--is the the one into which you've already substituted.
JamesJ
  • JamesJ
Hence you have: .02 = 0(.005) + .5(a)(.005)^2 This already simplifies to \[ \frac{0.005^2}{2}a = 0.02 \] Now solve for \( a \).
anonymous
  • anonymous
A = .02(2/.005^2) = .04/.005^2?
JamesJ
  • JamesJ
Evaluate it. You can't leave the answer in that form.
anonymous
  • anonymous
1600?
anonymous
  • anonymous
What about the units though, is that in meters?
JamesJ
  • JamesJ
We were (or at least I was) careful to convert all the units into SI units before we started calculating time = seconds distance = meters velocity = meters/sec = m/s acceleration = m/s^2
anonymous
  • anonymous
Ah, so its 1.6 km/s^2?
JamesJ
  • JamesJ
yes
anonymous
  • anonymous
Awesome! Thanks for all your help!

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