An electron has an initial speed of 178000 m/s. If it undergoes an acceleration of 4.8 x 1014 m/s2, how far has it traveled before reaching a speed of 789000 m/s? Answer in units of m.
Why are you posting this twice? We do not provide instant answers.
To find the distance traveled by the electron before reaching a speed of 789,000 m/s, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity = 789,000 m/s
u = initial velocity = 178,000 m/s
a = acceleration = 4.8 x 10^14 m/s^2
s = distance traveled
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Plugging in the given values, we get:
s = (789,000^2 - 178,000^2) / (2 * 4.8 x 10^14)
s = (622,521,000,000 - 31,684,000,000) / (9.6 x 10^14)
s = 590,837,000,000 / (9.6 x 10^14)
s ≈ 6.14 x 10^-4 m (rounded to 3 decimal places)
Therefore, the electron has traveled approximately 0.000614 m before reaching a speed of 789,000 m/s.
To find the distance traveled by the electron, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (789000 m/s)
u = initial velocity (178000 m/s)
a = acceleration (4.8 x 10^14 m/s^2)
s = distance traveled
Rearranging the equation to solve for s, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values into the equation:
s = (789000^2 - 178000^2) / (2 * 4.8 x 10^14)
Let's calculate this:
s = (622521000000 - 31684000000) / (9.6 x 10^14)
s = 590837000000 / (9.6 x 10^14)
s ≈ 0.614 m
Therefore, the electron has traveled approximately 0.614 meters before reaching a speed of 789000 m/s.