Hope Roth

2023-03-22

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?

Hussut8tv

Beginner2023-03-23Added 8 answers

Step 1: Given

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

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, $\u2206t=\frac{v-u}{a}\left(\because a=\frac{\u2206v}{\u2206t}\right)$

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, $t=\frac{10\times {10}^{4}-5\times {10}^{4}}{{10}^{4}}=5s$

Step 4: Calculate the distance covered in the meantime

By using the first equation, the distance covered is

$s=5\times {10}^{4}\times 5+\frac{1}{2}\times {10}^{4}\times {5}^{2}\therefore s=37.5\times {10}^{4}m$

Hence,

It takes time to double the starting velocity of $5s$.Distance covered while doubling the initial velocity is $37.5m$.

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

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, $\u2206t=\frac{v-u}{a}\left(\because a=\frac{\u2206v}{\u2206t}\right)$

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, $t=\frac{10\times {10}^{4}-5\times {10}^{4}}{{10}^{4}}=5s$

Step 4: Calculate the distance covered in the meantime

By using the first equation, the distance covered is

$s=5\times {10}^{4}\times 5+\frac{1}{2}\times {10}^{4}\times {5}^{2}\therefore s=37.5\times {10}^{4}m$

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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