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