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An electron moving with a velocity of $5 \times 10^{4}\, ms ^{-1}$ enters into a uniform electric 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 ?
$20 \,s$ and $7.5 \times 10^{4}\, m$
$5 \,s$ and $37.5 \times 10^{4}\, m$
$0.5 \,s$ and $75.3 \times 10^{4}\, m$
$15 \,s$ and $35.7 \times 10^{4}\, m$
Solution
Given initial velocity, $u=5 \times 10^{4} \,ms ^{-1}$
And acceleration, $a=10^{4} \,ms ^{-2}$
$(i)$ final velocity $=v=2 u=2 \times 5 \times 10^{4} \,ms ^{-1}=10 \times 10^{4}\, ms ^{-1}$
To find $t,$ use $v=u+a t$
Or $t=\frac{v-u}{a}$
$\left(\frac{10 \times 10^{4}-5 \times 10^{4}}{10^{4}}\right)=\frac{5 \times 10^{4}}{10^{4}}=5 \,s$
$(ii)$ Using $s=u t+\frac{1}{2} a t^{2}$
$=\left(5 \times 10^{4}\right) \times 5+\frac{1}{2}\left(10^{4}\right) \times(5)^{2}=25 \times 10^{4}+\frac{25}{2} \times 10^{4}=37.5 \times 10^{4}\, m$
