A subway train starting from rest leaves a
station with a constant acceleration. At the end of 5.59 s, it is moving at 11.5154 m/s.
What is the train’s displacement in the first 4.15337 s of motion?
Answer in units of m
Well, let me calculate that for you using my trusty clown calculator.
Given that the train starts from rest, has a constant acceleration, and reaches a velocity of 11.5154 m/s after 5.59 seconds, we can use the equation of motion:
v = u + at
where:
v = final velocity = 11.5154 m/s
u = initial velocity = 0 m/s (since it started from rest)
a = acceleration
t = time = 5.59 s
Solving for acceleration, we have:
a = (v - u) / t
a = (11.5154 - 0) / 5.59
a ≈ 2.064 m/s²
Now, to find the displacement in the first 4.15337 seconds, we can use the equation of motion:
s = ut + (1/2)at²
where:
s = displacement
u = initial velocity = 0 m/s
t = time = 4.15337 s
a = acceleration ≈ 2.064 m/s²
Plugging in the values, we get:
s = 0 + (1/2)(2.064)(4.15337)²
s ≈ 18.5 m
So, the train's displacement in the first 4.15337 seconds of motion is approximately 18.5 meters.
To find the displacement of the train in the first 4.15337 s of motion, we can use the equation:
displacement = initial velocity * time + (1/2) * acceleration * time^2
Since the train starts from rest, the initial velocity is 0 m/s.
Plugging in the given values:
displacement = 0 * 4.15337 + (1/2) * acceleration * (4.15337)^2
We need to find the acceleration of the train. To do that, we can use the formula:
final velocity = initial velocity + acceleration * time
Plugging in the given values:
11.5154 = 0 + acceleration * 5.59
Simplifying the equation:
11.5154 = 5.59 * acceleration
Dividing both sides by 5.59:
acceleration = 11.5154 / 5.59
Now we can substitute the value of acceleration back into the initial equation:
displacement = 0 * 4.15337 + (1/2) * (11.5154 / 5.59) * (4.15337)^2
Calculating the final answer:
displacement = (1/2) * (11.5154 / 5.59) * 17.249906
displacement ≈ 20.384 m
Therefore, the train's displacement in the first 4.15337 s of motion is approximately 20.384 m.
To find the displacement of the train during the first 4.15337 seconds of motion, we can use the equation of motion:
s = ut + (1/2)at²
where:
s = displacement
u = initial velocity (0 m/s since the train starts from rest)
t = time interval (4.15337 s)
a = constant acceleration
Given:
u = 0 m/s
t = 4.15337 s
Now, we need to find the acceleration of the train. We can use the equation:
v = u + at
where:
v = final velocity (11.5154 m/s)
u = initial velocity (0 m/s)
a = acceleration
Given:
v = 11.5154 m/s
u = 0 m/s
Rearranging the equation, we get:
a = (v - u) / t
Substituting the given values,
a = (11.5154 m/s - 0 m/s) / 5.59 s
a = 2.0639 m/s²
Now we have the acceleration. Plugging the values into the displacement equation:
s = (0 m/s)(4.15337 s) + (1/2)(2.0639 m/s²)(4.15337 s)²
s = 0 m + (1/2)(2.0639 m/s²)(17.252047 s²)
s = 0 m + 2.12695615 m²/s² * 17.252047 s²
s = 2.12695615 m²/s² * 17.252047 s²
s = 36.6825 m²/s²
Therefore, the train's displacement in the first 4.15337 seconds of motion is approximately 36.6825 meters.
The acceleration rate is
a = 11.5154/5.59 = 2.06 m/s^2
Displacement at t = 4.15337s =
(1/2) a t^2 = 17.77 m