Hope Roth

2023-03-22

An electron moving with a velocity of $5×{10}^{4}{\mathrm{ms}}^{-1}$ enters into a uniform electric field and acquires a uniform acceleration of ${10}^{4}{\mathrm{ms}}^{-2}$ 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?

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Step 1: Given
The electron's initial velocity, $u=5×{10}^{4}{\mathrm{ms}}^{-1}$
Acceleration acquired by the electron, $a={10}^{4}{\mathrm{ms}}^{-2}$
Step 2: Formulas used
The three motion equations tell us that,
Distance covered, $s=ut+\frac{1}{2}a{t}^{2}$,
Where u represents the starting speed., $t$ is the time taken and $a$ is the acceleration of the body.
We also know that,
time, $∆t=\frac{v-u}{a}\left(\because a=\frac{∆v}{∆t}\right)$
Step 3: Calculate the time required to double the initial velocity
We need that the final velocity is twice the initial velocity, thus, $v=2u=2×5×{10}^{4}={10}^{5}{\mathrm{ms}}^{-1}$.
Thus, the time required to acquire double the initial velocity is, $t=\frac{10×{10}^{4}-5×{10}^{4}}{{10}^{4}}=5s$
Step 4: Calculate the distance covered in the meantime
By using the first equation, the distance covered is
$s=5×{10}^{4}×5+\frac{1}{2}×{10}^{4}×{5}^{2}\therefore s=37.5×{10}^{4}m$
Hence,
It takes time to double the starting velocity of $5s$.Distance covered while doubling the initial velocity is $37.5m$.

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