Fm=Fe and am>ae
Hopefully this helps
As per kinematics equation we are given that

now we are given that
a = 2.55 m/s^2


now we need to find x
from above equation we have



so it will cover a distance of 93.2 m
Answer:
3 cm
Explanation:
According to the question,
.
.
.
Now the approximate slit's image width is equal to width of central maxima.
And width of central maxima is twice the width from center to first maxima
So,
.
Substitute all the variable in above equation.
.
.
In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s