Question

Please help me It's urgent​

Answer 1

$\huge\sf\underline{\underline{Given:-}}$

An electron is moving with velocity 5 × 10⁴ m/s .

It acquires a uniform acceleration of 10⁴ m/s² in its initial direction .

$\rule{10cm}{0.0356cm}$

$\huge\sf\underline{\underline{To \: Find :- }}$

• $Time \: taken \: for \: doubling \: its \: initial \: velocity$
• $Distance \: travelled$

$\rule{10cm}{0.0356cm}$

$\huge\sf\underline{\underline{Formulas \: Used:-}}$

$\boxed{\sf a \: = \frac{v - u}{t} }$

$\boxed{s \: = ut \: + \frac{1}{2} at {}^{2}}$

$\rule{10cm}{0.0356cm}$

$\large\sf\boxed{\underline{\underline{ANSWER }}}$

STEP 1

$\large Find \: Time$

$\boxed{\sf\red{a = \frac{v - u}{t} }}$

$\sf t \: = \frac{v - u}{a}$

$\sf u \: = 5 \times {10}^{4} ms {}^{ - 1}$

$\sf v \: = 10 \times {10}^{4} ms {}^{ - 1}$

$\sf a \: = {1 0}^{4} ms {}^{ - 2}$

$\sf t \: = \frac{(10 - 5) \times {10 }^{4} }{ {10}^{4} }$

$\sf \: \: \: \: = 10 - 5$

$\sf\blue{ \: \: \: \: = 5s} \:$

$\rule{10cm}{0.0356cm}$

STEP 2

$Find \: Distance$

$\boxed{\sf\red{ s = ut + \frac{1}{2} at²}}$

$\sf\implies s \: = 5 \times {10}^{4} \times 5 + (\frac{1}{2} \times ( {10}^{4} \times 5 {}^{2} ))$

$\sf\implies s \: \: \: \: = 25 \times {10}^{4} + \frac{( {10}^{4} \times 5 {}^{2} )}{2}$

$\sf\pink{ \: \: \: \: \: = 3.75 \times {10}^{5} m} \:$

$\rule{10cm}{0.0356cm}$

• $\sf\pink{Time \: taken = 5 second}$

• $\sf\orange{Distance \: travelled = 3.75 × 10⁵ m}$