Question

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 ${10}^4{\mathrm{ms}}^{-2}$ 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?

Answer 1

Step 1: Given

The initial velocity of the electron, $u=5\times {10}^4{\mathrm{ms}}^{-1}$

Acceleration acquired by the electron, $a={10}^4{\mathrm{ms}}^{-2}$

Step 2: Formulas used

From the three equations of motion, we know that,

Distance covered, $s=ut+\frac{1}{2}a{t}^2$,

Where $u$ is the initial velocity, $t$ 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, $v=2u=2\times 5\times {10}^4={10}^5{\mathrm{ms}}^{-1}$.

Thus, the time required to acquire double the initial velocity is, $t=\frac{10\times {10}^4-5\times {10}^4}{{10}^4}=5\mathrm{s}$

Step 4: Calculate the distance covered in the meantime

By using the first equation, the distance covered is

$\begin{array}{rcl}s& =& 5\times {10}^4\times 5+\frac{1}{2}\times {10}^4\times 5^2\\ \therefore s& =& 37.5\times {10}^4\mathrm{m}\end{array}$

Therefore,

1. Time required to double the initial velocity is $5\mathrm{s}$.
2. Distance covered while doubling the initial velocity is $37.5\mathrm{m}$.