Question

An electron moving with a velocity of 5 × 104 ms⁻¹ enters into a uniform electric field and acquires a uniform acceleration of 104 ms^–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 at this time?

Answer 1

Here initial velocity of the electron,

u = 5 × 10⁴ ms⁻¹ 

acceleration, a = 10⁴  ms⁻² 

(i) v = 2u = 2 × 5 × 10⁴ ms⁻¹ 

= 10 ×10⁴ ms

using v = u + at we get

10 × 10⁴ = 5 × 10⁴ + 10⁴ × t or t = 5s

(ii) Using s = `"ut" + 1/2 "at"^2`, we get

s = `5 xx 10^4 xx 5 + 1/2 xx 10^4 xx 5^2`

= 25 × 10⁴ + 12.5 × 10⁴ 

= 37.5 × 10⁴ m.