Question

A region is 1.5 cm in length and the electrons enter with a speed of 1x10^5 m/s and leaves with a speed of 2.5x10^6 m/s
What is their acceleration over this 1.5 cm length?

Answer 1

Using the equation of motion


displacement, initial velocity and final velocity are given, you can now find the acceleration

Answer 2

The equation of motion? Doesn't that need time?

Answer 3

isn't there any equation of motion in which there is no time?

Answer 4

Hi,
I think u guys have missed the equation
v^2 = u^2 + 2aS

where
v = final velocity
u = initial velocity
a = acceleration
S = Distance/Displacement

In ur case
u = 1x10^5 m/s
v = 2.5x10^6 m/s
S = 1.5 cm = 0.015 m
a = it's the question

Now all we have to do is just substituting the values in the equation

v^2 = u^2 +2aS
(2.5x10^6)^2 = (1x10^5)^2 + (2*0.015*a)
(2*0.015*a) = 10^10[ ( 6.25*10^2 - 1]
(2*0.015*a) = 624*10^10
a = 624*10^10/(2*0.015)
= 312*10^10 / (1.5*10^-2)
= 208 * 10^12 m/s^2

That's the answer for ur question
acceleration = 208 * 10^12 m/s^2

Hope this will help u !!!

Answer 5

I tried that answer but it said it was wrong? Is there another possible way to do it?

Answer 6

Oh wait never mind thanks for the help :) how do you determine how long it's been in the accelerating region?

Answer 7

Did u say "it said it was wrong" ?
Then it seems that u know something or someone who knows the correct answer. It would be really helpful if u could reply the answer 'cause then I could find out where I have messed up

If u want to find the time period of acceleration
S = ( u + v ) / t
(where the signs stand as same in the above one)

1.5 * 10^-2 = [(1x10^5) + (2.5x10^6)]/2 * t

t = 1.5/13*10^7 s

and u know
u could substitute it in
S = ut + 1/2at^2
and find "a"
but for my amazement it still gives me the same answer

Answer 8

Ya it's an computer that needs the correct answer but it was just a matter of plugging it in correctly, your answer is right :)