An electron moving with a velocity of `5 xx 10^(4)ms^(-1)` enters into a uniform electirc field and acquires a uniform acceleration of `10^(4)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 in this time ?

Explanation

Given , initial velocity, u = `5xx10^(4)ms^(-1)`
and acceleration , a = `10^(4)ms^(-2)`
(i) According to the question, final velocity, v = 2u, t = ?
From first equation of motion , v = u + at
`2u = u + 10^(4) xx t` [`because` put `v = 2u, a = 10^(4)ms^(-2)]`
`2u-u=10^(4)xxt`
`10^(4)xxt = u`
`rArr t=((u))/(10^(4))=(5xx10^(4))/(10^(4))=5s [because u = 5 xx10^(4)m//s]`
i.e., after 5 s electron will acquire a velocity double os its initial velocitt.
(ii) From second equation of motion.
Distance covered in t second, s = `ut + 1/2at^(2)`
`=5xx10^(4)xx5+1/2xx10^(4)(5)^(2)[put, u=5xx10^(4)m//s, t = 5s and a = 10^(4) m//s^(2)]`
=`25xx10^(4)+1/2xx10^(4)xx25`
=`25xx10^(4)+12.5xx10^(4)`
= `10^(4)(25+12.5)`
= `37.5xx10^(4)m=37.5xx10xx10^(3)m=375xx10^(3)m`
= 375 km [`because` 1000m=1km]