A subway train starting from rest leaves a station with a constant acceleration. At the end of 5.24 s, it is moving at 13.4144 m/s.
What is the train’s displacement in the first 3.42696 s of motion?
To find the train's displacement in the first 3.42696 seconds of motion, we can use the kinematic equation:
displacement = initial velocity * time + (1/2) * acceleration * time^2
We are given the following information:
- The train starts from rest, meaning the initial velocity is 0 m/s.
- The time is 3.42696 s.
- We need to find the displacement.
Since the train starts from rest, the initial velocity is 0. Therefore, the equation simplifies to:
displacement = (1/2) * acceleration * time^2
Now, let's substitute the known values into the equation:
displacement = (1/2) * acceleration * (3.42696 s)^2
To find the acceleration, we can use the equation:
final velocity = initial velocity + acceleration * time
We are given the following additional information:
- At the end of 5.24 s, the train's velocity is 13.4144 m/s.
- The initial velocity is 0 m/s.
- The time is 5.24 s.
Using these values, we can rearrange the equation to solve for acceleration:
acceleration = (final velocity - initial velocity) / time
Substituting the known values:
acceleration = (13.4144 m/s - 0 m/s) / 5.24 s
Now we can substitute the calculated value of acceleration back into the displacement equation:
displacement = (1/2) * [(13.4144 m/s - 0 m/s) / 5.24 s] * (3.42696 s)^2
Evaluating this expression will give us the displacement in the first 3.42696 seconds of motion.
a = (Vf-Vo)/t,
a = (13.4144-0)/5.24 = 2.56 m/s^2.
D = Vo*t + 0.5a*t^2,
D = 0 + 1.28*(3.42696)^2 = 15.03 m.