Question

A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s\(^2\) for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50 m/s\(^2\) until it stops at the next station. Find the total distance covered.

Short answer

The total distance covered by the train is 1796.48 meters.

Step-by-step solution

Calculate acceleration distance

Using the formula for distance covered under constant acceleration, \( s = ut + \frac{1}{2} a t^2 \), where \( u = 0 \) (initial velocity), \( a = 1.60 \text{ m/s}^2 \), and \( t = 14.0 \text{ s} \), we calculate the distance: \[s = 0 + \frac{1}{2} \times 1.60 \times (14)^2 = 0.80 \times 196 = 156.8 \text{ m}. \]This is the distance covered during acceleration.

Calculate constant speed

After accelerating, the train reaches a constant speed. We find this speed using the formula \( v = u + at \). Since \( u = 0 \text{ m/s} \), \( a = 1.60 \text{ m/s}^2 \), and \( t = 14.0 \text{ s} \), it becomes:\[v = 0 + 1.60 \times 14 = 22.4 \text{ m/s}. \]This is the speed at which the train travels during the next phase.

Calculate distance traveled at constant speed

Using the distance formula for constant speed \( s = v \times t \), where \( v = 22.4 \text{ m/s} \) and \( t = 70.0 \text{ s} \), we find the distance:\[s = 22.4 \times 70 = 1568 \text{ m}.\]This is the distance covered while traveling at constant speed.

Calculate deceleration distance

Using the formula \( s = ut + \frac{1}{2} a t^2 \) for deceleration, where \( v = 0 \) (final velocity), \( a = -3.50 \text{ m/s}^2 \), and using, \( v^2 = u^2 + 2as \) to find \( s \), initially:\[0 = (22.4)^2 + 2(-3.50)s \to s = \frac{(22.4)^2}{2 \times 3.50} = \frac{501.76}{7} = 71.68 \text{ m}.\]This is the distance over which the train decelerates.

Calculate total distance covered

Add the distances from all three phases: acceleration, constant speed, and deceleration. \[\text{Total distance} = 156.8 + 1568 + 71.68 = 1796.48 \text{ m}.\]This is the total distance covered by the train.

Key concepts

Constant Acceleration

When an object speeds up steadily, it is said to be under constant acceleration. In the given exercise, our subway train starts from rest and accelerates at a constant rate of 1.60 m/sy^2 for a time of 14.0 seconds. This constant acceleration means that every second, the train's velocity increases by 1.60 m/s.
To calculate the distance covered during this period of constant acceleration, we use the formula for distance: \[ s = ut + \frac{1}{2} a t^2 \] Given that the initial velocity \( u \) is zero, the effect is straightforward—the distance increases proportionally to the square of the time and the rate of acceleration. This formula is essential for any scenario involving motion in a straight line with constant acceleration.
You could visualize this phase as a segment of time where the train picks up speed smoothly and steadily. The total distance during this phase was calculated to be 156.8 meters.

Deceleration

Deceleration is similar to acceleration but involves slowing down instead of speeding up. It's also known as negative acceleration. The train undergoes deceleration as it approaches the next station. Here, the rate of deceleration is 3.50 m/sy^2.
The main point about deceleration is that it results in a decrease in velocity over time until the train stops. In our calculation, it was crucial to understand that despite the train moving initially at a speed of 22.4 m/s, deceleration reduced this speed to zero.
To find the distance the train travels while decelerating, we use the equation: \[ v^2 = u^2 + 2as \] Rearranging it allows us to solve for \( s \), giving us the distance during deceleration as 71.68 meters. Just like with acceleration, this formula helps us understand the dynamics of slowing down effectively.

Distance Covered

Calculating the total distance covered involves adding the distances traversed during each phase of motion: acceleration, constant speed, and deceleration. We've already calculated each separate distance based on the motion characteristics specific to each phase.

• Acceleration Distance: 156.8 meters
• Constant Speed Distance: 1568 meters
• Deceleration Distance: 71.68 meters

Adding these values provides the total path traveled by the subway train: 1796.48 meters. Each segment of the journey contributes to this figure, giving a complete picture of motion from start to finish.

The understanding of how to break down the journey into manageable parts is crucial for solving problems in physics related to motion.

Velocity Calculation

Velocity is a key concept in kinematics. It differs from speed in that it's a vector, which means it has both magnitude and direction. In the problem, calculating the velocity during acceleration was necessary to determine how fast the train travels as it transitions from rest.
Using the equation of motion: \[ v = u + at \] we find the final velocity after acceleration. This calculation showed that the velocity reached by the train was 22.4 m/s after 14 seconds of constant acceleration. This velocity became the train's speed during its constant speed phase.
Velocity calculations aren't just limited to the acceleration phase but play a role in determining how motion progresses into the next stages. Understanding velocity empowers one to predict subsequent motion patterns effectively.