An electron moving with a velocity of $5 \times 10^{4} \mathrm{~ms}^{-1}$ enters into a uniform electric field and acquires a uniform acceleration of $104 \mathrm{~ms}^{-2}$ 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?
Here, $u=5 \times 10^{4} \mathrm{~ms}^{-1}, a=10^{4} \mathrm{~ms}^{-2}$
(i) Let after time $t, v=2 u=10 \times 10^{4} \mathrm{~ms}^{-1}$
Using $v=u+a t$, we get
$10 \times 10^{4}=5 \times 10^{4}+10^{4} t$ or $10^{4} t=5 \times 10^{4}$
$\therefore t=5 \mathrm{~s}$
(ii) Distance Covered by electron in $5 s$
$\mathrm{S}=u t+\frac{1}{2} a t^{2}=5 \times 10^{4} \times 5+\frac{1}{2} \times 10^{4} \times 25$
$=25 \times 10^{4}+12 \cdot 5 \times 10^{4}=37 \cdot 5 \times 10^{4} \mathrm{~m}=375 \mathrm{~km}$
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