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 $10^4\ m s^{-2}$ 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 of electron $u=5\times 10^4\ ms^{-1}$

Acceleration $a=10^4\ m s^{-2}$

To do: $(i)$. To calculate the time in which the electron would acquire a velocity double its initial velocity.

$(ii)$. To know how the distance is covered by the electron at this time.

Solution: 

$(i)$. As the electron would acquire a velocity double its initial velocity. So, its final velocity $v=2u=2\times5\times 10^4\ ms^{-1}=10^5\ ms^{-1}$

On using the first equation of motion, $v=u+at$

$10^5\ ms^{-1}=5\times 10^4\ ms^{-1}+10^4\ m s^{-2}\times t$

Or $10^5-5\times10^4=10^4\ m s^{-2}\times t$

Or $5\times10^4=10^4t$

Or $t=\frac{5\times10^4}{10^4}$

Or $t=5\ second$

Therefore, the electron would acquire a velocity double its initial velocity in $5\ seconds$.

$(ii)$. On using second equation of the motion, $s=ut+\frac{1}{2}at^2$

$s=5\times10^4\times5+\frac{1}{2}\times 10^4\ m s^{-2}\times 5^2$

Or $s=25\times10^4+12.5\times10^4$

Or $s=37.5\times10^4\ m$

Therefore, the distance covered by the electron at this time is $37.5\times10^4\ m$.

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Updated on: 10-Oct-2022

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