An electron, starting from rest and moving
with a constant acceleration, travels 5.5 cm in
10 ms.
What is the magnitude of this acceleration?
Answer in units of km/s
Vf = d/t=5.5*10^-2m / 10*10^-3s=5.5m/s
a = (VF - Vo) / t,
a = (5.5 - 0) / 0.01 = 550m/s^2.
a = 550m/s^2 * (1/1000)km/m=0.55km/s^2.
To find the magnitude of acceleration, we can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Given:
distance = 5.5 cm = 0.055 m
time = 10 ms = 0.01 s
initial velocity = 0 m/s (starting from rest)
Rearranging the formula to solve for acceleration:
acceleration = (distance - (initial velocity * time)) / (0.5 * time^2)
Substituting the given values into the formula:
acceleration = (0.055 m - (0 m/s * 0.01 s)) / (0.5 * (0.01 s)^2)
acceleration = (0.055 m) / (0.00005 s^2)
acceleration โ 1100 m/s^2
To convert from m/s^2 to km/s^2, we divide by 1000:
acceleration โ 1.1 km/s^2
Therefore, the magnitude of the acceleration is 1.1 km/s^2.
To find the magnitude of acceleration, we can use the kinematic equation:
\[d = v_0 t + \frac{1}{2} a t^2\]
where:
- \(d\) is the displacement (5.5 cm in this case)
- \(v_0\) is the initial velocity (0 m/s in this case, as the electron starts from rest)
- \(a\) is the acceleration (unknown)
- \(t\) is the time taken (10 ms, which is equivalent to 0.01 s)
Rearranging the equation to solve for \(a\), we have:
\[a = \frac{2(d - v_0 t)}{t^2}\]
Substituting the given values, we get:
\[a = \frac{2(0.055 m - 0 \cdot 0.01 s)}{(0.01 s)^2}\]
Simplifying the equation, we have:
\[a = \frac{0.11 m}{0.0001 s^2} = 1100 m/s^2\]
Since we need to answer in units of km/s, we convert meters to kilometers by dividing by 1000, and convert seconds to hours by multiplying by 3600:
\[a = \frac{1100 m/s^2}{1000 \cdot 3600 s/h}\]
Simplifying further:
\[a = 0.000305 km/s^2\]
Therefore, the magnitude of the acceleration of the electron is 0.000305 km/s^2.