A subway train starting from rest leaves a station with a constant acceleration. At the end of 7.94s, it is moving at 12.3864m/s. What is the train's displacement in the first 5.57388s of motion?
Answer in unites of m.
To find the train's displacement in the first 5.57388 seconds of motion, we can use the equation for displacement with constant acceleration:
\(s = u t + \frac{1}{2} a t^2\)
where:
s is the displacement
u is the initial velocity (starting from rest, u = 0)
t is the time
a is the acceleration
We are given:
u = 0 m/s (starting from rest)
t = 5.57388 s
a = ?
To find the acceleration, we can rearrange the equation for final velocity with constant acceleration:
\(v = u + a t\)
Given that v = 12.3864 m/s and t = 7.94 s, we can solve for a:
\(a = \frac{v - u}{t} = \frac{12.3864 - 0}{7.94}\)
Now that we have the acceleration, we can substitute the values into the first equation to find the displacement:
\(s = (0)(5.57388) + \frac{1}{2} \left(\frac{12.3864 - 0}{7.94}\right) (5.57388)^2\)