1) The distance travelled by the electron is [tex]1\cdot 10^{17} m[/tex]
2) The time taken is [tex]4.0\cdot 10^{10}s[/tex]
Explanation:
1)
The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance travelled
For the electron in this problem,
[tex]u=5\cdot 10^6 m/s[/tex] is the initial velocity
v = 0 is the final velocity (it comes to a stop)
[tex]a=-1.25\cdot 10^{-4} m/s^2[/tex] is the acceleration
Solving for s, we find the distance travelled:
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m[/tex]
2)
The total time taken for the electron in its motion can also be found by using another suvat equation:
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
Here we have
[tex]u=5\cdot 10^6 m/s[/tex]
v = 0
[tex]a=-1.25\cdot 10^{-4} m/s^2[/tex]
And solving for t, we find the time taken:
[tex]t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s[/tex]
Learn more about accelerated motion:
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brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
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