Electron Velocity from Linearly Increasing Acceleration Graph

[emyt]

Homework Statement

an electron starting from rest has an acceleration that increases linearly with time, a= kt.
k = 1.5ms^2/s

Plot acceleration v.s. time graph for 10 seconds

Estimate electron's velocity 5.0s after its motion starts

Homework Equations

vx =vx0 + axt

The Attempt at a Solution

a = 1.5(5) = 7.5

v = 7.5(5) = 37.5

the velocity at 5 seconds is 37.5 m/s?

thanks

 

[Vykan12]

You used a kinematics formula that only applies if acceleration is constant. You have to integrate the given expression for acceleration to get the correct one for velocity.

You should get something like v(t) = v_{0} + \frac{1}{2}kt^2

 

[emyt]

You used a kinematics formula that only applies if acceleration is constant. You have to integrate the given expression for acceleration to get the correct one for velocity.

You should get something like v(t) = v_{0} + \frac{1}{2}kt^2

thanks, that's strange that equation isn't in my book - how do you come up with it?

thanks again

 

[Vykan12]

There's no equation for it because there's infinitely many ways that acceleration can vary as a function of time.

Eg/

a(t) = \sin(t)

v(t) = \int_{0}^{t} \sin(t) \, dt = -\cos(t) + C

 

[emyt]

is there a way to "estimate" the velocity after 5 seconds without integration? I'm pretty sure we aren't expected to use this method 2 chapters into the book

thanks

 

[Vykan12]

You could draw a graph and find the area of the triangle formed by the line, the y-axis and some line x=a.

 

[emyt]

You could draw a graph and find the area of the triangle formed by the line, the y-axis and some line x=a.

okay, to plot a velocity graph, do I just take the acceleration values at each point and use those as slopes?

so if the acceleration at 5 seconds is 7.5 m/s^2 then ill have a 7.5x + c tangent line?

thanks

 

[emyt]

You could draw a graph and find the area of the triangle formed by the line, the y-axis and some line x=a.

I am not sure what you mean.. if you wanted to find the acceleration at a specific point, how do you draw a triangle?

thanks

 

[emyt]

hello, I'm waiting :( I sincerely need help, thanks

 

[Vykan12]

Do you understand why the area of a rectangle in a graph of acceleration versus time is an estimation of instantaneous velocity?

Recall that for constant acceleration we have a = v/t so v= at.

I really can't explain this properly without the use of calculus. Maybe someone else can.

 

[emyt]

Do you understand why the area of a rectangle in a graph of acceleration versus time is an estimation of instantaneous velocity?

Recall that for constant acceleration we have a = v/t so v= at.

I really can't explain this properly without the use of calculus. Maybe someone else can.

yes, because it's acceleration times time

okay, so do I take the area of a super small interval of the graph?

thanks

 

[Vykan12]

See the attachment I just posted.

Edit: Sorry, t should be 5 in that graph and a should be (1.5)(5) = 7.5.

 

[emyt]

See the attachment I just posted.

Edit: Sorry, t should be 5 in that graph and a should be (1.5)(5) = 7.5.

thanks so much, I'll look at it when it's finished pending approval

 

[emyt]

how long does it take for a picture to pend approval? :|

 

[Vykan12]

Forget the attachment. I'll use image shack from now on.

http://img34.imageshack.us/img34/4853/areatriangle.jpg

 

[emyt]

hi, I see that you got 7.5, how did you get 7.5 = 18.75?

thanks

 

[Vykan12]

It should say (0.5)(5)(7.5) = 18.75. The equation got cut into 2 lines by accident.

 

[emyt]

It should say (0.5)(5)(7.5) = 18.75. The equation got cut into 2 lines by accident.

:S weird.. I actually tried that before but I couldn't get the right answer, I guess I miscalculated something and threw out the whole plan... the constant acceleration formula works because its at the instant of 5 seconds right?

and I can plot the velocity versus time graph by just doing various points..et c?

thanks a lot

 

[Vykan12]

You're wrong on both counts. The constant acceleration formula cannot be used unless the graph of a versus t is a horizontal line. As for the v versus t graph, you need to compute the area from 0 to t_{1}, then from 0 to t_{2}, then from 0 to t_{3} and so on to plot various points v(1), v(2), v(3), etc.

Again, just wait for these kinematics topics to be covered in more detail in class or your book or wherever you're learning from.

 

[emyt]

You're wrong on both counts. The constant acceleration formula cannot be used unless the graph of a versus t is a horizontal line. As for the v versus t graph, you need to compute the area from 0 to t_{1}, then from 0 to t_{2}, then from 0 to t_{3} and so on to plot various points v(1), v(2), v(3), etc.

Again, just wait for these kinematics topics to be covered in more detail in class or your book or wherever you're learning from.

"the constant acceleration formula cannot be used unless the graph of a versus t is a horizontal line. " oh right, you just did the area under the curve, I mistook that as the v = at formula
" As for the v versus t graph, you need to compute the area from 0 to t_{1}, then from 0 to t_{2}, then from 0 to t_{3} and so on to plot various points v(1), v(2), v(3), etc."

yeah that's what I meant by doing various points.. at least partially, since I was thinking about the wrong formula anyway

thanks a bunch! :)