A train starts from rest and accelerates uniformly, until it has traveled 3000 m and acquired a velocity of 24 m/s. The train then moves a constant velocity of 24 m/s for 430 s. The train then decelerates uniformly at 0.065 m/s², until it is brought to a halt.
1. In Situation A, the acceleration during the first 3.0 km of travel is closest to
2. In Situation A, the distance traveled by the train during deceleration, in kilometers is closest to
3. In Situation A, the velocity of the train, when it has decelerated for 160 seconds, is closest to
4.In Situation A, the average velocity, during the first 8.0 km of travel, is closest to
Use your tried-and-true equations:
v = at
s = 1/2 at^2
#1.
3000 = 1/2 at^2
t = √(6000/a)
24 = a√(6000/a) = √6000a
a = 24^2/6000 = 0.096 m/s^2
and so on
To solve this problem, we can use the basic equations of motion. Let's break down each part of the train's motion and calculate the required quantities.
1. To find the acceleration during the first 3.0 km of travel, we first need to calculate the time it takes for the train to reach a velocity of 24 m/s. We can use the equation:
v = u + at
Where:
v is the final velocity (24 m/s)
u is the initial velocity (0 m/s)
a is the acceleration (unknown)
t is the time
Plugging in the values, we get:
24 = 0 + a * t
Since the train starts from rest (0 m/s), the equation simplifies to:
24 = a * t
Next, we can calculate the time taken to reach a velocity of 24 m/s using the equation:
s = ut + (1/2) * a * t^2
Where:
s is the displacement (3000 m)
u is the initial velocity (0 m/s)
a is the acceleration (unknown)
t is the time
Plugging in the values, we get:
3000 = (1/2) * a * t^2
Simplifying the equation, we get:
6000 = a * t^2
Now, we can substitute the value of t from the first equation into the second equation:
6000 = a * (24/a)^2
Simplifying further, we get:
6000 = 24^2
Solving for a:
a = (24^2) / 6000
Using a calculator, we find:
a ≈ 0.096 m/s²
Therefore, the acceleration during the first 3.0 km of travel is closest to 0.096 m/s².
2. To find the distance traveled by the train during deceleration, we need to calculate the time it takes for the train to come to a halt. We can use the equation:
v = u + at
Where:
v is the final velocity (0 m/s, since the train comes to a halt)
u is the initial velocity (24 m/s)
a is the acceleration (-0.065 m/s², deceleration is negative)
t is the time
Plugging in the values, we get:
0 = 24 + (-0.065) * t
Solving for t:
24 = 0.065 * t
t ≈ 369.23 s
Now, we can calculate the distance traveled during the deceleration phase using the equation:
s = ut + (1/2) * a * t^2
Where:
s is the displacement (unknown)
u is the initial velocity (24 m/s)
a is the acceleration (-0.065 m/s², deceleration is negative)
t is the time (369.23 s)
Plugging in the values, we get:
s = 24 * 369.23 + (1/2) * (-0.065) * (369.23)^2
Using a calculator, we find:
s ≈ -2671681.94 m
Since distance cannot be negative, the distance traveled during deceleration is approximately 0 km.
3. To find the velocity of the train when it has decelerated for 160 seconds, we can use the equation:
v = u + at
Where:
v is the final velocity (unknown)
u is the initial velocity (24 m/s)
a is the acceleration (-0.065 m/s², deceleration is negative)
t is the time (160 s)
Plugging in the values, we get:
v = 24 + (-0.065) * 160
Using a calculator, we find:
v ≈ 13 m/s
Therefore, the velocity of the train, when it has decelerated for 160 seconds, is closest to 13 m/s.
4. To find the average velocity during the first 8.0 km of travel, we can use the equation:
v_avg = (u + v) / 2
Where:
v_avg is the average velocity (unknown)
u is the initial velocity (0 m/s)
v is the final velocity (unknown)
Since we know the initial and final velocities are the same (both 24 m/s), we can simplify the equation:
v_avg = (24 + 24) / 2
Using a calculator, we find:
v_avg = 24 m/s
Therefore, the average velocity during the first 8.0 km of travel is closest to 24 m/s.