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Found in: Page 33

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# An electron with an initial velocity${{\mathbf{\text{V}}}}_{{\mathbf{0}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{50}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$enters a region of length L=1.00cm where it is electrically accelerated. It emerges with${\mathbf{v}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{70}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{6}}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$. What is its acceleration, assumed constant?

The acceleration of electron to be assumed constant is $1.62×{10}^{15}\mathrm{m}/{\mathrm{s}}^{2}$ .

See the step by step solution

## Step 1: Given Data

Initial velocity, ${\mathrm{v}}_{0}=1.50×{10}^{5}\mathrm{m}/\mathrm{s}$

Final velocity, $\mathrm{v}=5.70×{10}^{5}\mathrm{m}/\mathrm{s}$

Distance travelled, x=1.00 cm

## Step 2: Understanding the concept of acceleration

Acceleration can be found by using the given data in the kinematic equation which involves initial velocity, final velocity, acceleration, and displacement terms.

The formula to find acceleration can be obtained with the help of final as well as initial velocity.

${v}^{2}={v}_{0}^{2}+2ax\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathrm{v}\mathrm{is}\mathrm{the}\mathrm{final}\mathrm{velocity},{\mathrm{v}}_{0}\mathrm{is}\mathrm{the}\mathrm{initial}\mathrm{velocity},\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{acceleration}\mathrm{and}\mathrm{x}\mathrm{is}\mathrm{the}\mathrm{distance}\mathrm{travelled}.$

## Step 3: Determination of the acceleration.

Convert 1.00 cm into i.e. in S.I. units.

$1\mathrm{cm}=\frac{1\mathrm{cm}}{100\mathrm{cm}}×1\mathrm{m}\phantom{\rule{0ex}{0ex}}=0.01m\phantom{\rule{0ex}{0ex}}\mathrm{Rearrange}\mathrm{the}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{acceleration}\mathrm{as},\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{{\mathrm{v}}^{2}-{\mathrm{v}}_{0}^{2}}{2\mathrm{x}}\phantom{\rule{0ex}{0ex}}\mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}.\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{{\left(5.7×{10}^{6}\mathrm{m}/\mathrm{s}\right)}^{2}-{\left(1.5×{10}^{5}\mathrm{m}/\mathrm{s}\right)}^{2}}{2×0.01\mathrm{m}}\phantom{\rule{0ex}{0ex}}=1.62×{10}^{15}\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{electron}\mathrm{to}\mathrm{be}\mathrm{assumed}\mathrm{constant}\mathrm{is}1.62×{10}^{15}\mathrm{m}/{\mathrm{s}}^{2}.$