Question

An electron moving with a velocity of 5 × 104 m s -1 enters into a uniform electric field and acquires a uniform acceleration of 104 m s -2 in the direction of its initial motion. (i) Calculate the time in which the electron would acquire a velocity double of its initial velocity. (ii) How much distance the electron would cover in this tim

Answer 1

(i) t = 5s

(ii) x = 3.75*10^5 m

Step-by-step explanation:

(i) To calculate the time that the electron takes to reach twice the value of its initial velocity, you use the following formula:

`v=v_o+at` (1)

vo: initial velocity of the electron = 5*10^4 m/s

v: final velocity of the electron = 2vo = 1*10^5 m/s

a: acceleration of the electron = 1*10^4 m/s^2

You solve the equation (1) for t, and replace the values of the parameters:

`t=(v-v_o)/(a)=(1*10^5m/s-5*10^4m/s)/(1*10^4m/s^2)=5s`

The electron takes 5s to reach twice its initial velocity.

(ii) The distance traveled by the electron in such a time is:

`x=v_ot+(1)/(2)at^2` (2)

you replace the values of the parameters in the equation (2):

`x=(5*10^4m/s)(5s)+(1)/(2)(1*10^4m/s^2)(5s)^2\\\\x=3.75*10^5m`

The distance traveled by the electron is 3.75*10^3m/s