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 time?

given

Initial velocity u = 5x104 m s-1


And acceleration a = 104 m s-2


(i) According to the given question, the final velocity v = 2u


We need to find the time, t =?


V = u+ at


2u = u +(104 ms-2 ) × t


u = 104 ms-1 × t


t = u / 104 m s-1


t = (5x104) / 104


=5s


(ii) t=5s, a=104 ms-2, u=5x104ms-1 s=?


s=ut + at2


s = (5 x104) x5 +1/2(104) x (5)2


s = 25 x104 + 25/2 x (104)


s = 25 x 104 +12.5 x (104)


s = 37.5x104 m


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