A light–rail commuter train draws 645 A of 590–V DC electricity when accelerating. What is its power consumption rate in kilowatts?
How long does it take to reach 30.0 m/s starting from rest if its loaded mass is 4.50×104 kg, assuming 91.0% efficiency and constant power?
Find its average acceleration.
What is the ratio of the acceleration you found for the light–rail train compares to what might be typical for an automobile? (A typical automobile can accelerate from 0 to 60 mph in 10 s.)
645 * 590 = 380,550 W or 381 kW
Ke = (1/2)m v^2 = (1/2)(4.5*10^4)(900)
= 2025*10^4 = 2.025*10^7 Joules
energy needed = (2.025/.91)10^7
= 2.225 * 10^7 Joules
3.8055*10^5 t = 2.225*10^7
t = .585*10^2 = 58.5 seconds, about a minute
a = delta v/delta t = 30/58.5 = .513 m/s^2
approx 60* (50 km/h/30 mi/h) = 100 km/h
100/3.6 = 28 m/s
28 m/s / 10 s = 2.8 m/s^2
car accelerates 2.8/.513 = 5.5 times as fast
To find the power consumption rate of the light-rail commuter train, we can use the formula: Power (P) = Current (I) * Voltage (V).
1. First, we need to convert the current from amperes (A) to kiloamperes (kA). Since 1 kiloampere is equal to 1000 amperes, we divide the current by 1000: 645 A / 1000 = 0.645 kA.
2. Now we can calculate the power consumption rate by multiplying the current and voltage: P = 0.645 kA * 590 V = 380.55 kW.
Therefore, the power consumption rate of the light-rail commuter train is 380.55 kilowatts.
To find the time it takes for the light-rail train to reach 30.0 m/s, we can use the formula: Time (t) = (2 * Distance (d)) / (Efficiency (η) * Power (P)).
1. The distance (d) is not provided in the question. If it is given, you can substitute it directly into the formula.
2. The efficiency (η) is given as 91.0%. To convert it to a decimal, divide it by 100: 91.0% / 100 = 0.91.
3. The power (P) is already calculated as 380.55 kW.
4. Substitute all the values into the formula: t = (2 * d) / (0.91 * 380.55 kW).
Since the distance is not given, it is not possible to determine the exact time it takes for the train to reach 30.0 m/s without additional information.
To find the average acceleration of the light-rail train, we can rearrange the formula: Power (P) = Force (F) * velocity (v) and substitute mass (m) for force.
1. Rearrange the formula to solve for force: F = P / v.
2. The velocity (v) is given as 30.0 m/s.
3. The power (P) is calculated as 380.55 kW. To convert it to watts, multiply by 1000: 380.55 kW * 1000 = 380550 W.
4. Substitute the values into the formula: F = 380550 W / 30.0 m/s.
The force can then be used to find the average acceleration using Newton's second law, F = m * a, where m is the mass.
1. The mass (m) is given as 4.50 * 10^4 kg.
2. Rearrange the formula to solve for acceleration: a = F / m.
3. Substitute the values into the formula: a = (380550 W / 30.0 m/s) / (4.50 * 10^4 kg).
The average acceleration of the light-rail train can be calculated using the values from the previous steps.
To compare the acceleration of the light-rail train to a typical automobile, we can calculate the acceleration of the automobile first.
1. Convert the acceleration of an automobile from mph to m/s. 60 mph = 26.82 m/s.
2. The time it takes for the automobile to accelerate from 0 to 60 mph is given as 10 seconds.
3. Calculate the average acceleration of the automobile using the formula: a = (final velocity - initial velocity) / time.
a = (26.82 m/s - 0 m/s) / 10 s = 2.682 m/s^2.
4. Divide the average acceleration of the light-rail train calculated earlier by the average acceleration of the automobile:
Ratio = (acceleration of light-rail train) / (acceleration of automobile).
The ratio will give you an indication of how the acceleration of the light-rail train compares to a typical automobile.
To find the power consumption rate in kilowatts, we can use the formula: P = IV, where P is the power in watts, I is the current in amps, and V is the voltage in volts.
Given:
Current (I) = 645 A
Voltage (V) = 590 V
To convert the power from watts to kilowatts, divide the power by 1000.
Step 1: Calculate the power consumption rate
P = IV
P = 645 A * 590 V
P = 380550 W
Step 2: Convert watts to kilowatts
Power consumption rate = P / 1000
Power consumption rate = 380550 W / 1000
Power consumption rate = 380.55 kW
Therefore, the power consumption rate of the light-rail commuter train is 380.55 kilowatts.
To calculate the time it takes for the train to reach 30.0 m/s starting from rest, we can use the equation:
\( t = \frac{v_f - v_i}{P \cdot \eta} \)
where t is the time, P is the power, η is the efficiency, v_f is the final velocity, and v_i is the initial velocity.
Given:
Loaded mass (m) = 4.50 × 10^4 kg
Efficiency (η) = 91.0%
Final velocity (v_f) = 30.0 m/s
Initial velocity (v_i) = 0
Step 1: Calculate the power
Power (P) = Power consumption rate (380.55 kW)
Step 2: Calculate the time
t = (v_f - v_i) / (P * η)
t = (30.0 m/s - 0) / (380550 W * 0.91)
Step 3: Solve for t
t = 30.0 m/s / (380550 W * 0.91)
Therefore, the time it takes for the train to reach 30.0 m/s starting from rest is t = 0.092 seconds.
To find the average acceleration, we can use the formula:
\( a = \frac{v_f - v_i}{t} \)
Given:
Final velocity (v_f) = 30.0 m/s
Initial velocity (v_i) = 0
Time (t) = 0.092 s
Step 1: Calculate the average acceleration
a = (v_f - v_i) / t
a = (30.0 m/s - 0) / 0.092 s
Therefore, the average acceleration of the light-rail train is a = 326.086 m/s^2.
To find the ratio of the acceleration of the light-rail train to a typical automobile, we need to convert the automobile's acceleration from mph to m/s.
Given:
Auto acceleration = 0 to 60 mph in 10 s
Step 1: Convert mph to m/s
1 mph = 0.44704 m/s
Auto acceleration = (60 mph) * (0.44704 m/s per mph) / 10 s
Auto acceleration = 2.68224 m/s^2
Step 2: Calculate the ratio of the two accelerations
Ratio = (acceleration of light-rail train) / (acceleration of automobile)
Ratio = 326.086 m/s^2 / 2.68224 m/s^2
Therefore, the ratio of the acceleration of the light-rail train to a typical automobile is approximately 121.5.