# An electron with initial velocity v0 = 1.10 × 10^5 m/s enters a region 1.0 cm long where it is electrically accelerated (see the figure). It emerges with velocity v = 5.37 × 10^6 m/s. What is its acceleration, assumed constant?

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I want someone to double check my answer

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Gió Share
Sep 7, 2015

I found: $a = 1.44 \times {10}^{15} \frac{m}{s} ^ 2$

#### Explanation:

I am going to use non-relativistic kinematics concepts here although the veleocities are quite high.

You can use the relationships:
${v}_{f} = {v}_{0} + a t$
${x}_{f} - {x}_{i} = d = {v}_{0} t + \frac{1}{2} a {t}^{2}$
where:
$d = 1 c m = 0.01 m$ is the distance (linear) described;
$v$ is velocity;
$a$ is acceleration.
From the first we get:
$t = \frac{{v}_{f} - {v}_{0}}{a} = \frac{5.37 \times {10}^{6} - 1.10 \times {10}^{5}}{a} = \frac{5.26 \times {10}^{6}}{a}$
to simplfy let us set: $\frac{5.26 \times {10}^{6}}{a} = \frac{k}{a}$ and substitute into the second:
$0.01 = 1.10 \times {10}^{5} \cdot \frac{k}{a} + \frac{1}{2} \cancel{a} {k}^{2} / {a}^{\cancel{2}}$
rearranging and using our values:
$0.01 a = 1.10 \times {10}^{5} \cdot 5.26 \times {10}^{6} + \frac{1}{2} {\left(5.26 \times {10}^{6}\right)}^{2}$
$a = 1.44 \times {10}^{15} \frac{m}{s} ^ 2$

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